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JEE MAIN 2020 - 02 SEP S2 2020

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CHEMISTRY25 questions

Q1JEE Main 2020MCQ4MOrganic Chemistry - Some Basic Principles

The major product of the following reaction is:

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Q2JEE Main 2020MCQ4MIonic Equilibrium

If you spill a chemical toilet cleaning liquid onyour hand, your first aid would be :

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📖 Explanation

Toilet cleaning liquid has about $10.5 \%$ w/v ${HCl} ;$ to neutralise its affect aqueous ${NaHCO}_{3}$ is used while ${NaOH}$ is avoid for this purposebecause its highly corosive in nature and canburn body.

Q3JEE Main 2020MCQ4MOrganic Chemistry - Some Basic Principles

Arrange the following labelled hydrogens in decreasing order of acidity :

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📖 Explanation

Acidic strength order : $R-C^{∥}^{O}-OH>R-OH$ $>R-C\equiv CH$ Reason : $R-C^{∥}^{O}-O^{⊖}$ stable by equivalent resonance. StableSo answer is ${b}> {c}> {d}> {a}$ .

Q4JEE Main 2020MCQ4Md and f Block Elements

Cast iron is used for the manufacture of :

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Q5JEE Main 2020MCQ4MAlcohols, Phenols and Ethers

Two compounds A and B with same molecular formula $({C}_{3} {H}_{6} {O} )$ undergo Grignard's reaction with methylmagnesium bromide to give products ${C}$ and ${D}$ . products ${C}$ and ${D}$ show following chemical tests. Test C D Cericammoniumnitrate Test Positive Positive Lucas Test Turbidity obtained after five minutes Turbidity obtained immediately Iodoform Test Positive NegativeC and D respectively are :

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Q6JEE Main 2020MCQ4MChemical Bonding and Molecular Structure

The shape/structure of $[{XeF}_{5} ]^{-}$ and ${XeO}_{3} {F}_{2}$ respectively, are :

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📖 Explanation

$\text{table} {XeF}_{5} ^{-}, {XeO}_{3} {F}_{2}; {sp}^{3} {d}^{3},{sp}^{3} {d} ;\text{Pentagonal} \text{planar} ,\text{Trigonal} \text{bipyramidal}$

Q7JEE Main 2020MCQ4MHaloalkanes and Haloarenes

The major product obtained from $\ {E}_{2}$ -elimination of 3 -bromo- 2 -fluoropentaneis:

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Q8JEE Main 2020MCQ4MClassification of Elements

Three elements $\ {X}, \ {Y}$ and $\ {Z}$ are in the $3 ^ {\text{rd}}$ period of the periodic table. The oxides of $\ {X}, \ {Y}$ and Z, respectively, are basic, amphoteric and acidic. The correct order of the atomic numbers of $\ {X}, \ {Y}$ and $\ {Z}$ is :

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📖 Explanation

When we are moving from left to right in a periodic table acidic character of oxides increases (as well as atomic number of atom increases) $\therefore \ {X}< {Y}< {Z} (\text{ acidic character })$ ${X}< {Y}< {Z} \text{(atomic number) }$

Q9JEE Main 2020MCQ4MStructure of Atom

The number of subshells associated with $\ {n}=4$ and $\ {m}=-2$ quantum numbers is :

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📖 Explanation

For $\ {n}=4$ $l=0,1,2,3$ $\therefore 4 \ {d} \&$ 4f subshell associated with $\ {n}=4, \ {m}=-2$

Q10JEE Main 2020MCQ4MCoordination Compounds

Simplified absorption spectra of three complexes ((i), (ii) and (iii)) of ${M}^{ {n}+}$ ion are provided below; their $\lambda_{\max }$ values are marked as ${A}, {B}$ and ${C}$ respectively. The correct match between the complexes and their $\lambda_{\max }$ values is :(i) $[{M}( {NCS})_{6 }]^{(-6+ {n})}$ (ii) $[{MF}_{6} ]^{(-6+ {n})}$ (iii) $[{M}({NH}_{3} )_{6} ]^{ {n}+}$

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📖 Explanation

Stronger the ligand greater is splitting of d orbitals and smaller will be wavelength of light absorbed. The splitting of ligand is ${F}^{-}< {NCS}^{-}< {NH}_{3}$ So order of wavelength of light absorbed is $(\lambda _{\max })_{ {NH}_{3}} < (\lambda _{\max })_{ {NCS}^{-}}< (\lambda _{\max })_{ {F}^{-}}$

Q11JEE Main 2020MCQ4MChemical Kinetics

Consider the reaction sequence given below :Which of the following statements is true:

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📖 Explanation

Reaction $1: \ {SN}_{1}$ Reaction $2: \ {E}_{2}$ $\ {SN}_{1}$ is independent of concentration of nucleophile/base

Q12JEE Main 2020MCQ4MChemical Kinetics

The results given in the below table were obtained during kinetic studies of the following reaction: $2 \ {A}+\ {B} \to {C}+\ {D}$ Experiment [A]/molL $^{-1}$ [B]/molL $^{-1}$ Initialrate/molL $^{-1}$ min $^{-1}$ I 0.1 0.1 $6.00 \times 10^{-3}$ II 0.1 0.2 $2.40 \times 10^{-2}$ III 0.2 0.1 $1.20 \times 10^{-2}$ IV X 0.2 $7.20 \times 10^{-2}$ V 0.3 Y $2.88 \times 10^{-1}$ $\ {X}$ and $\ {Y}$ in the given table are respectively:

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📖 Explanation

From rate law ${r}=- \frac{1}{2}\frac{{d}[ {A}]}{ {dt}}= \frac{- {d}[ {B}]}{ {dt}}$ $= {K}[ {A}]^{ {x}}[ {B}]^{ {y}}$ $6 \\times 10^{-3}=\ {K}(0.1)^{\ {x}}(0.1) \ {y}$ .........(1) $2.4 \\times 10^{-2}=\ {K}(0.1)^{\ {x}}(0.2) \ {y}$ .......(2) $1.2 \\times 10^{-2}=\ {K}(0.2)^{\ {x}}(0.1) \ {y}$ ........(3) $(3) \÷(1) \Rightarrow {x}=1$ $(2) \÷(3) \Rightarrow {x}=2$ So, other with respect to $\ {A}=1$ Order with respect to $\ {B}=2$ $(4) \÷(3)$ $(\frac{{x}}{0.2} )\times (\frac{0.2}{0.1} )^{2}$ $= \frac{7.2 \\times 10^{-2}}{1.2 \\times 10^{-2}}$ $x=\ \frac{6 \\times 0.2}{4}$ $x=0.3 \ {M}$ $(5) \÷(4)$ $(\frac{y}{0.2} )^{2}= \frac{2.88 \\times 10^{-1}}{7.2 \\times 10^{-2}}$ $y^{2}=4 \\times 0.2^{2}$ $y=0.4 \ {M}$

Q13JEE Main 2020MCQ4MAldehydes, Ketones and Carboxylic Acids

An organic compound 'A' $( {C}_{9} {H}_{10} {O} )$ whentreated with conc. HI undergoes cleavage toyield compounds 'B' and 'C'. 'B' gives yellowprecipitate with ${AgNO}_{3}$ where as 'C' tautomerizes to 'D'. 'D' gives positive idoform test. 'A' could be :

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Q14JEE Main 2020MCQ4MSolutions

The size of a raw mango shrinks to a muchsmaller size when kept in a concentrated saltsolution. Which one of the following processescan explain this?

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Q15JEE Main 2020MCQ4MClassification of Elements

Two elements A and B have similar chemicalproperties. They don't form solidhydrogencarbonates, but react with nitrogen toform nitrides. A and B, respectively, are :

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📖 Explanation

Both Li and Mg form nitride when reactsdirectly with nitrogen. The hydrogen carbonate of both $\ {Li}$ and $\ {Mg}$ does not exist in solid state. All alkali metal hydrogen carbonate exist in solid state except ${LiHCO}_{3}$ .

Q16JEE Main 2020MCQ4MCoordination Compounds

The one that is not expected to show isomerismis :

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📖 Explanation

$[Ni(NH_3 ) _2Cl _2 ]$ is tetrahedral complex, therefore does not show geometrical and optical isomerism. $[Ni(NH_3 ) _2Cl _2 ]$ does not show structural isomerism $[{Ni}({NH}_{3} )_{4}({H}_{2} {O} )_{2} ]^{2+} ] \ &$ $[Pt(NH _3 ) _2Cl _2 ]$ show geometrical isomerism $[Ni(en)_3 ] ^{2+}$ show optical isomerism

Q17JEE Main 2020MCQ4MChemical Equilibrium

Amongst the following statements regarding adsorption, those that are valid are : (a) $\Delta{H}$ becomes less negative as adsorptionproceeds. (b) On a given adsorbent, ammonia isadsorbed more than nitrogen gas. (c) On adsorption, the residual force actingalong the surface of the adsorbentincreases. (d) With increase in temperature, theequilibrium concentration of adsorbateincreases.

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📖 Explanation

(a) since adsorption is exothermic process, asadsorption proceeds number of active sitespresent over adsorbent decreases, so less heatis evolved. (b) since $\ {NH}_{3}$ has higher force of attraction onadsorbent due to its polar nature (high value of 'a'). (c) As the adsorption increases, residual forces oversurface decreases. (d) Since process is exothermic, on increasingtemperature it shift to backward direction, soconcentration of adsorbate particle decreases.

Q18JEE Main 2020MCQ4MChemical Bonding and Molecular Structure

Match the type of interaction in Column A with the distance dependence of their interaction energy in Column B： A B (I) iron - ion (a) $1/r$ (II) dipole - dipole (b) $1/r^2$ (III) London dispersion (c) $1/r^3$ (a) $1/r^6$

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📖 Explanation

Type of interaction Interaction Energy(E) ion - ion ${E} \propto\frac{1}{ {r}}$ dipole - dipole ${E} \propto\frac{1}{ {r}^3}$ London dispersion ${E} \propto\frac{1}{ {r}^6}$

Q19JEE Main 2020MCQ4MBiomolecules

The correct observation in the following reactions is : Sucrose ${\to }^{\text{Glycosidic} \text{bond}}_{\text{Cleavage}(\text{Hydrolysis})}$ A + B ${\to }^{\text{Seliwanoff} 's}_{ (\text{reagent})}$

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📖 Explanation

Seliwanoff's test is used to distinguish between aldose and ketone sugars; when added to a solution containing ketose, red colour is formed rapidly. Sucrose ${\to }^{\text{Hydrolysis}}$ Glucose + Fructose ${\to }^{\text{Seliwanoff} 's}_{ (\text{reagent})}$ Redcolour

Q20JEE Main 2020MCQ4MChemical Bonding and Molecular Structure

The molecular geometry of $SF_6$ is octahedral. What is the geometry of $SF_4$ (including lone pair(s) of electrons, if any) ?

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📖 Explanation

$4\sigma$ bonds +1 lone pair ∴ Shape (including lone pair of electrons) is Trigonal bipyramidal

Q21JEE Main 2020NAT4MChemical Thermodynamics

The heat of combustion of ethanol into carbondioxides and water is -327 kcal at constantpressure. The heat evolved (in cal) at constantvolume and $27°{C}$ (if all gases behave ideally)is $(R=2$ cal ${mol}^{-1} {K}^{-1} )$

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📖 Explanation

${C}_{2} {H}_{5} {OH}_{(l)}+3 {O}_{2( {g})} { \to }$ $2 {CO}_{2( {g})}+3 {H}_{2} {O}_{(l)}$ $\Delta{n}_{ {g}}=2-3=-1$ $\Delta_{ {c}} {H}=\Delta_{ {c}} {U}+ (\Delta{n}_{ {g}} ) {RT}$ $\Delta_{ {c}} {H}=\Delta_{ {c}} {U}- {RT}$ $\Delta_{ {c}} {U}=\Delta_{ {c}} {H}+ {RT}$ $=-327 \\times 10^{3}+2 \\times 300$ $=-326400 \ {cal}$ $\therefore$ Heat evolved $=326400$ cal.

Q22JEE Main 2020NAT4MElectrochemistry

For the disproportionation reaction $2 {Cu}^{+}( {aq})⇌{Cu}( {s})+ {Cu}^{2+}(\ {aq})$ at $298 \ {K}$ ln $\ {K}$ (where $\ {K}$ is the equilibrium constant) is _____ $\times 10^{-1}$ Given $({E}°_{ {Cu}^{2+} / {Cu}^{+}} =0.16 {V},$ $\ {E}°_{ {Cu}^{+} / {Cu}} =0.52 {V},$ $\frac{{RT}}{ {F}}=0.025 )$

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📖 Explanation

$\ {Cu}^{+} { ─── ▸}{Cu}+ {e}^{-}$ $\ {Cu}^{+} +e^{-} { ─── ▸}{Cu} (s)$ ────────────── $2 {Cu}^{+} { ─── ▸} {Cu}^{2+} +{Cu}$ ${E}°_{ \text{cell}}={E}°_{ {Cu}^{2+} / {Cu}^{+}}-{E}°_{ {Cu}^{+} / {Cu}}$ $=0.52-0.16$ $=0.36 \ {V}$ At equilibrium $\to {E}_{\text{cell }}=0$ ${E}°_{\text{cell }} =\frac{{RT}}{ {nF}} \ln {K}$ $\ln {K}={{E}°_{\text{cell }}\times {nF}}/{ {RT}}$ $\ln {K}= \frac{0.36 \\times 1}{0.025}$ $=14.4=144 \\times 10^{-1}$

Q23JEE Main 2020NAT4Md and f Block Elements

The oxidation states of transition metal atoms in ${K}_{2} {Cr}_{2} {O}_{7}, {KMnO}_{4}$ and ${K}_{2} {FeO}_{4},$ respectively, are ${x}, {y}$ and ${z} .$ The sum of ${x}, {y}$ and ${z}$ is _____.

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📖 Explanation

${K}_{2} {Cr}_{2} {O}_{7}$ $2 (+1) + 2x + 7(-2) = 0$ In $K_2Cr_2O_7$ , Transition metal (Cr) present in +6 oxidation state. $KMnO_4$ $(+1) + y + 4(-2) = 0$ $x = +7$ In ${KMnO}_{4},$ transition metal (Mn) present in +7 oxidation state ${K}_{2} {FeO}_{4}$ $2(+1)+z+4(-2)=0$ $x=+6$ In ${K}_{2} {FeO}_{4},$ transition metal (Fe) present in +6oxidation state So, $x=+6$ $y=+7$ $z=+6$ ───────── $x+y+z=19$

Q24JEE Main 2020NAT4MOrganic Chemistry - Some Basic Principles

The ratio of the mass percentages of 'C & H' and 'C $\&$ O' of a saturated acyclic organiccompound 'X' are 4: 1 and 3: 4 respectively. Then,the moles of oxygen gas required for completecombustion of two moles of organic compound'X' is _____.

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📖 Explanation

C : H = 4 : 1 C : O = 3 : 4 Mass ratio C : H : O = 12 : 3 : 16 Mole ratio C : H : O = 1 : 3 : 1 Empirical formula = $CH_3 O$ Molecular formula $= C_2 H_6 O_2$ (saturated acyclic organic compound) $C_2 H_6 O _2 + 5/2 O _2 \to 2CO_2$ $+ 3H_2 O$ $2 \text{mole} \;\;\;\;\;\;5 \text{mol}$ Moles of $O_2$ required = 5 moles

Q25JEE Main 2020NAT4MStructure of Atom

The work function of sodium metal is $4.41 \\times 10^{-19} {J} .$ If the photons of wavelength $300 \ {nm}$ are incident on the metal, the kinetic energy of the ejected electrons will be $(\ {h}=6.63 \\times 10^{-34} {J} {s} ;$ ${c}=3 \\times 10^{8} \ {m} / \ {s}\ )$ _____ $\\times 10^{-21} \ {J}$

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📖 Explanation

${E}= {W}+ {K}.{E}_{\max }$ ${K} .{E}_{\max }= {E}- {W}$ $=\frac{{hc}}{\lambda}-4.41 \\times 10^{-19}$ $= \frac{6.63 \\times 10^{-34} \\times 3 \\times 10^{8}}{300 \\times 10^{-9}}$ $-4.41 \\times 10^{-19}$ $=2.22 \\times 10^{-19} \ {J}$ $=222 \\times 10^{-21} \ {J}$

MATHEMATICS25 questions

Q26JEE Main 2020MCQ4MParabola

The area (in sq. units) of an equilateral triangle inscribed in the parabola $\ {y}^{2}=8 \ {x},$ with one of its vertices on the vertex of this parabola, is :

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📖 Explanation

$\tan 30°= \frac{4 {t}}{2 {t}^{2}}= \frac{2}{ {t}}$ $\Rightarrow {t}=2 \\sqrt{3}$ ${AB}=8 {t}=16 \\sqrt{3}$ Area $=256.3 .\frac{\\sqrt{3}}{4}$ $=192 \\sqrt{3}$

Q27JEE Main 2020MCQ4MPermutations and Combinations

Let n > 2 be an integer. Suppose that there are n Metro stations in a city located along a circular path. Each pair of stations is connected by a straight track only. Further, each pair of nearest stations is connected by blue line, whereas all remaining pairs of stations are connected by red line. If the number of red lines is 99 times the number of blue lines, then the value of n is :-

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📖 Explanation

Number of blue lines $=$ Number of sides $=\ {n}$ Number of red lines $=$ number of diagonals $={ }^{n} C_{2}-n$ ${ }^{n} C_{2}-n=99 {n}$ $\Rightarrow \frac{n(n-1)}{2}-n=99 \ {n}$ $\ \frac{n-1}{2}-1=99$ $\Rightarrow n=201$

Q28JEE Main 2020MCQ4MTrigonometric Ratios and Identities

If the equation $\cos ^{4} \theta + sin \;^{4} \theta +\lambda =0$ has realsolutions for $\theta ,$ then $\lambda$ lies in the interval:

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📖 Explanation

$\lambda=- (\sin ^{4} \theta +\cos ^{4} \theta )$ $\lambda=- (\sin ^{2} \theta +\cos ^{2} \theta )^{2}$ $-2 \sin ^{2} \theta \cos ^{2} \theta$ $\lambda=\ \frac{\sin ^{2} 2 \theta }{2}-1$ $\frac{\sin ^{2} 2 \theta }{2} \in [0, \ \frac{1}{2} ]$ $\lambda\in [-1,- \frac{1}{2} ]$

Q29JEE Main 2020MCQ4MQuadratic Equation and Inequalities

Let $\ {f}( {x})$ be a quadratic polynomial such that ${f}(-1)+ {f}(2)=0 .$ If one of the roots of ${f}( {x})=0$ is $3,$ then its other root lies in :

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📖 Explanation

$f(x)=a(x-3)(x-\alpha )$ $f(2)=a(\alpha -2)$ $f(-1)=4 a(1+\alpha )$ $f(-1)+f(2)=0$ $\Rightarrow a(\alpha -2+4+4 \alpha )=0$ $a \ne 0 \Rightarrow5 \alpha =-2$ $\alpha =-\ \frac{2}{5}=-0.4$ $\alpha \in (-1,0)$

Q30JEE Main 2020MCQ4MSequences and Series

Let $\ {f}: \ {R}\to {R}$ be a function which satisfies $f(x+y)=f(x)+f(y) \forall x,$ $y \in R .$ If $f(1)=2$ and $\ {g}( {n})=∑_{\ {k}=1} ^{(\ {n}-1)} {f}( {k}), {n} \in {N}$ then the value of $\ {n},$ forwhich $g(n)=20,$ is :

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📖 Explanation

$f(x+y)=f(x)+f(y)$ $\Rightarrowf(n)=n f(1)$ $f(n)=2 n$ $g(n)=∑_{\ {k}=1} ^{ {n}-1 } 2 n$ $=2 ( \frac{(n-1) n}{2} )=n(n-1)$ $g(n)=20$ $\Rightarrow n(n-1)=20$ $n=5$

Q31JEE Main 2020MCQ4MMatrices and Determinants

Let ${a}, {b}, {c} \in {R}$ be all non-zero and satisfy $\ {a}^{3}+\ {b}^{3}+\ {c}^{3}=2 .$ If the matrix $A+(\text{table} a,b,c;b,c,a;c,a,b)$ satisfies ${A}^{{T}} {A}= {I},$ then a value of abc can be :

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📖 Explanation

$\ {A}^{ {T}} {A}=\ {I}$ $\Rightarrow {a}^{2}+ {b}^{2}+ {c}^{2}=1$ and ${ab}+ {bc}+ {ca}=0$ Now, $( {a}+ {b}+ {c})^{2}=1$ $\Rightarrow {a}+ {b}+ {c}=\± 1$ So $a^{3}+b^{3}+c^{3}-3 a b c$ $=( {a}+ {b}+ {c})$ $( {a}^{2}+ {b}^{2}+ {c}^{2}- {ab}-{bc}- {ca} )$ $=\±1(1-0)=\±1$ $\Rightarrow3 \ {abc}=2 \±1=3,1$ $\Rightarrow {abc}=1, \ \frac{1}{3}$

Q32JEE Main 2020MCQ4MApplication of Derivatives

Let $\ {f}:(-1, \infty ) \to {R}$ be defined by $\ {f}(0)=1$ and $\ {f}({x})= \frac{1}{ {x}} \log _{ {e}}(1+{x}), {x} \ne 0 .$ Then the function $\ {f}$

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📖 Explanation

$f'(x)=\frac{\frac{x}{1+x}-l n(1+x)}{x^{2}}$ $= \frac{x-(1+x) ln(1+x)}{x^{2}(1+x)}$ Suppose ${h}( {x})= {x}-(1+ {x}) l{n}(1+ {x})$ $\Rightarrow {h}'( {x})=1-\ln (1+ {x})-1$ $=-\ln (1+ {x})$ ${h}'( {x})>0, \forall {x} \in (-1,0)$ ${h}'( {x})<0, ∀{x} \in (0, \infty )$ ${h}(0)=0 \Rightarrow {h}'( {x})<0 ∀{x} \in (-1, \infty )$ $\Rightarrow {f}'( {x})<0∀{x} \in (-1, \infty )$ $\Rightarrow {f}( {x})$ is a decreasing function for all ${x} \in (-1$ $\infty )$

Q33JEE Main 2020MCQ4MSequences and Series

If the sum of first 11 terms of an A.P., $\ {a}_{1} \ {a}_{2}, \ {a}_{3}, \....$ is $0 (a_{1} \ne 0\ ),$ then the sum of the A.P., $\ {a}_{1}, \ {a}_{3}, \ {a}_{5},...., \ {a}_{23}$ is $\ {ka}_{1},$ where $\ {k}$ is equal to :

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📖 Explanation

$a_{1}+a_{2}+a_{3}+\.....+a_{11}=0$ $\Rightarrow (a_{1}+a_{11} ) \times \frac{11}{2}=0$ $\Rightarrow {a}_{1}+\ {a}_{11}=0$ $\Rightarrow {a}_{1}+\ {a}_{1}+10 \ {d}=0$ where d is common difference $\Rightarrow [a_{1}=-5 \ {d}]$ $a_{1}+a_{3}+a_{5}+.....+a_{23}$ $= (a_{1}+a_{23} ) \times \frac{12}{2}$ $= (a_{1}+a_{1}+22 d ) \\times 6$ $= (2 \ {a}_{1}+22(\frac{-\ {a}_{1}}{5})) \\times 6$ $=-\ \frac{72}{5} {a}_{1} \Rightarrow{K}=\ \frac{-72}{5}$

Q34JEE Main 2020MCQ4MComplex Numbers

The imaginary part of $(3+2 \\sqrt{-54})^{1 / 2}-(3-2 \\sqrt{-54})^{1 / 2}$ can be :

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📖 Explanation

$|3+2 \sqrt{-54}|=\sqrt{9+216}=15$ $\Rightarrow (3+2 \sqrt{-54})^{1 / 2}=±(\sqrt{\;\frac{15+3}{2}}+i \sqrt{\;\frac{15-3}{2}})$ $=±(3+i \sqrt{6})$ and $(3-2 \sqrt{-54})^{1 / 2}=±(3-i \sqrt{6})$ Hence $\{(3+2 \sqrt{-54})^{1 / 2}-(3-2 \sqrt{-54})^{1 / 2}\}$ $=± 2 i \sqrt{6}$ or ±6 Hence imaginary part $=± 2 \sqrt{6}$

Q35JEE Main 2020MCQ4MLimits, Continuity and Differentiability

$\lim _{x \to 0} (\tan (\frac{\pi }{4}+x))^{1 / x}$ is equal to :

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📖 Explanation

$\lim _{x \to 0} \{\tan (\frac{\pi }{4}+x) \}^{1 / x}$ $=e^{{\lim _{x \to 0} \frac{1}{x}} \{\tan (\frac{\pi }{4}+x )-1 \}}$ $=\ {e ^{{\lim _{x \to 0}(\ \frac{1+\tan x-1+\tan x}{x(1-\tan x)} )}$ $=e^{{\lim _{x \to 0}\frac{2 \tan x}{x(1-\tan x)}$ $=e^{2}$

Q36JEE Main 2020MCQ4MApplication of Derivatives

The equation of the normal to the curve $\ {y}=(1+\ {x})^{2 \ {y}}+\cos ^{2}\ (\sin ^{-1} \ {x}\ )$ at $\ {x}=0$ is :

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📖 Explanation

Given equation of curve $y=(1+x)^{2 y}+\cos ^{2}\ (\sin ^{-1} x\ )$ at $[{x=0}]$ $\ {y}=(1+0)^{2 \ {y}}+\cos ^{2}\ (\sin ^{-1} 0\ )$ $y=1+1$ $\ [{y}=2]$ So we have to find the normal at (0,2) Now $y=e^{2 y \ln (1+x)}$ $+\cos ^{2}\ (\cos ^{-1} \\sqrt{1-x^{2}}\ )$ $y=e^{2 y \ln (1+x)}+\ (\\sqrt{1-x^{2}}\ )^{2}$ $y=e^{2 y \ln (1+x)}+\ (1-x^{2}\ )$ ......(1) Now differentiate w.r.t. x $\ {y}'= {e}^{2 {y} \ln (1+{x})}$ $[2 {y} .( \frac{1}{1+ {x}} ) +\ln (1+ {x}) \.2 {y}']$ $-2 {x}$ Put $x=0 \ \&\ y=2$ ${y}'= {e}^{2 \\times 2l \ {n} 1}$ $[2 \\times 2 ( \frac{1}{1+0} )+\ln (1+0) . 2 {y}']$ $-2 \\times 0$ $y'= {e}^{0}[4+0]-0$ ${y}'=4=$ slope of tangent to the curve so slope of normal to the curve $=- \frac{1}{4} \{ {m}_{1} {m}_{2}=-1 \}$ Hence equation of normal at (0,2) is $y-2=- \frac{1}{4}(x-0)$ $\Rightarrow4 y-8=-x$ $\Rightarrow [ x+4 y=8]$

Q37JEE Main 2020MCQ4MEllipse

For some $\theta \in (0, \frac{\pi }{2}\ ),$ if the eccentricity of thehyperbola, $x^{2}-y^{2} \sec ^{2} \theta =10$ is $\\sqrt{5}$ times theeccentricity of the ellipse, $x^{2} \sec ^{2} \theta +y^{2}=5,$ thenthe length of the latus rectum of the ellipse, is:

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📖 Explanation

Given $\theta \in (0, \ \frac{\pi }{2}\ )$ equation of hyperbola $\Rightarrow{x}^{2}-\ {y}^{2} \sec ^{2} \theta =10$ $\Rightarrow \frac{{x}^{2}}{10}-\frac{{y}^{2}}{10 \cos ^{2} \theta }=1$ Hence eccentricity of hyperbola $({e}_{ {H}} )=\\sqrt{1+ \frac{10 \cos ^{2} \theta }{10}}$ .....(1) $\{ {e}=\\sqrt{1+\frac{{b}^{2}}{ {a}^{2}}} \}$ Now equation of ellipse $\Rightarrow {x}^{2} \sec ^{2} \theta + {y}^{2}=5$ $\Rightarrow \frac{{x}^{2}}{5 \cos ^{2} \theta }+\frac{{y}^{2}}{5}=1$ $\{ {e}=\\sqrt{1-\frac{{a}^{2}}{ {b}^{2}}} \}$ Hence eccenticity of ellipse $(e_{\ {E}} )=\\sqrt{1- \frac{5 \cos ^{2} \theta }{5}}$ $({e}_{\ {E}} )=\\sqrt{1-\cos ^{2} \theta }$ $=|\sin \theta |=\sin \theta$ ...(2) $\{\ \because \theta \in (0, \frac{\pi }{2} ) \}$ given $\Rightarrow {e}_{ {H}}=\\sqrt{5} {e}_{ {e}}$ Hence $1+\cos ^{2} \theta =5 \sin ^{2} \theta$ $1+\cos ^{2} \theta =5 (1-\cos ^{2} \theta )$ $1+\cos ^{2} \theta =5-5 \cos ^{2} \theta$ $6 \cos ^{2} \theta =4$ $\cos ^{2} \theta = \frac{2}{3}$ ...(3) Now length of latus rectum of ellipse $= \frac{2 {a}^{2}}{ {b}}= \frac{10 \cos ^{2} \theta }{\\sqrt{5}}$ $=\ \frac{20}{3 \\sqrt{5}}=\ \frac{4 \\sqrt{5}}{3}$

Q38JEE Main 2020MCQ4MSets and Relations

Which of the following is a tautology?

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📖 Explanation

Option (1) is $∼ {p} \ ∧( {p} \vee {q}) \to {q}$ $\equiv (\sim {p} \wedge {p}) \vee (\sim {p} \ ∧{q}) \to {q}$ $\equiv {Cv}(\sim {p} \wedge {q})\to {q}$ $\equiv (\sim {p} \ ∧{q}) \to {q}$ $\equiv \sim (\sim {p} \ ∧{q}) \vee {q}$ $\equiv ( {p} \vee \sim {q}) \vee {q}$ $\equiv ( {p} \vee {q}) \vee (\sim {q} \vee {q})$ $\equiv ( {p} \vee {q}) \vee {t}$ so $\sim {p} \ ∧( {p} \vee {q}) \to {q}$ is a tautology

Q39JEE Main 2020MCQ4MThree Dimensional Geometry

A plane passing through the point (3,1,1) contains two lines whose direction ratios are $1, -2, 2$ and $2,3,-1$ respectively. If this plane also passes through the point $(\alpha ,-3,5),$ then $\alpha$ is equal to:

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📖 Explanation

Hence normal is $\perp ^r$ to both the lines so normal vector to the plane is $\vec{n}=(\hat{i}-2\hat{j}+2\hat{k})$ $\times (2\hat{i}+3\hat{j}-\hat{k})$ $\vec{n}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 2 & 3 & -1\end{vmatrix}$ $=\hat{i}(2-6)-\hat{j}(-1-4)+\hat{k}(3+4)$ $\vec{n}=-4\hat{i}+5\hat{j}+7\hat{k}$ Now equation of plane passing through (3,1,1) is $\Rightarrow -4(x-3)+5(y-1)+7(z-1)=0$ $\Rightarrow -4 x+12+5 y-5+7 z-7=0$ $\Rightarrow -4 x+5 y+7 z=0$ ...(1) Plane is also passing through $(\alpha ,-3,5)$ so this point satisfies the equation of plane so put in equation (1) $-4 \alpha +5 \times (-3)+7 \times (5)=0$ $\Rightarrow -4 \alpha -15+35=0$ $\Rightarrow \alpha =5$

Q40JEE Main 2020MCQ4MProbability

Let ${E}^{{C}}$ denote the complement of an event ${E}$ . Let ${E}_{1}, {E}_{2}$ and ${E}_{3}$ be any pairwise independent events with ${P}({E}_{1}) > 0$ and ${P}({E}_{1} ∩{E}_{2} ∩{E}_{3})=0.$ Then ${P}({{E}_{2}^{{C}}∩{E}_{3}^{{C}}} / {E}_{1})$ is equal to

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📖 Explanation

Given ${E}_{1}, {E}_{2}, {E}_{3}$ are pairwise indepedent events so ${P} ({E}_{1} ∩ {E}_{2} )= {P}({E}_{1} ) .{P}({E}_{2} )$ and ${P}({E}_{2} ∩{E}_{3} )= {P}({E}_{2} ).{P}({E}_{3} )$ and ${P}({E}_{3} ∩{E}_{1} )= {P}({E}_{3} ).{P}$ {E}{1} )& \ {P}({E}{1} ∩{E}{2}∩{E}{3} )=0 $Now$ {P}({\overline{{E}}{2}∩ \overline{{E}}{3}}/{ {E}{1}} )={{P}[{E}{1} ∩(\overline{{E}}{2} ∩ \overline{{E}}{3})]}/{ {P}({E}{1} )}=\frac{{P}({E}{1} )-[{P}({E}{1} ∩{E}{2} )+ {P}({E}{1} ∩{E}{3} )- {P}({E}{1}∩{E}{2} ∩{E}{3} ) ]}{ {P}({E}{1} )}=\frac{{P}({E}{1} )- {P}({E}{1} ) .{P}({E}{2} )- {P}({E}{1} ) {P}({E}{3} )-0}{ {P}({E}{1} )}=1- {P}({E}{2} )- {P} ({E}{3} )= [1- {P}({E}{3} ) ]- {P}({E}{2} )={P}({E}{3}^{{C}})-{P}({E}{2})$

Q41JEE Main 2020MCQ4MMatrices and Determinants

Let ${A}= \{ {X}=( {x}, {y}, {z})^{ {T}}: {PX}=0$ $^{2}+y^{2}+z^{2}=1\ \}$ where $P=\begin{bmatrix}1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1\end{bmatrix}$ andthen the set A:

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📖 Explanation

Given $P=\begin{bmatrix}1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1\end{bmatrix}$ , Here $| {P}|=0 \ \&$ also given PX = 0 \Rightarrow $\begin{bmatrix}1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1\end{bmatrix}$ $\begin{bmatrix}x \\ y \\ z\end{bmatrix}=0$ \Rightarrow $\text{table} x+2 y+z=0;-2 x+3 y-4 z=0;x+9 y-z=0\}$ D = 0, so system have infinite many solutions, By solving these equation we get $x= \frac{-11 \lambda}{2} ; y=\lambda; z= \frac{7 \lambda}{2}$ Also given, $x^{2}+y^{2}+z^{2}=1$ $\Rightarrow ( \frac{-11 \lambda}{2} )^{2}+(\lambda)^{2}+ (\ \frac{7 \lambda}{2} )^{2}=1$ $\Rightarrow \lambda=±{1}/{\\sqrt{ \frac{121}{4}+1+\frac{49}{4}}}$ so, there are 2 values of $\lambda$ . $\therefore$ so, there are 2 solution set of $(\ {x}, {y}, \ {z})$

Q42JEE Main 2020MCQ4MArea Under The Curves

Consider a region ${R}= \{( {x}, {y}) \in {R}^{2}: {x}^{2} \le {y} \le 2 \ {x} \}$ If a line $\ {y}=\alpha$ divides the area of region $\ {R}$ into two equal parts, then which of the following is true?

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📖 Explanation

${y} \ge {x}^{2} \Rightarrow$ upper region of $\ {y}=\ {x}^{2}$ $\ {y} \le 2 \ {x} \Rightarrow$ lower region of $\ {y}=2 \ {x}$ According to ques, area of $\ {OABC}=2$ area of $\ {OAC}$ $\Rightarrow\int _{0} ^{4} (\\sqrt{ {y}}-\frac{{y}}{2} ) {dy}$ $=2 ∫_{0} ^{\alpha }(\\sqrt{\ {y}}-\frac{{y}}{2} ) .{dy}$ $\Rightarrow\ \frac{4}{3}=2[\frac{2}{3} \alpha ^{3 / 2}-\ \frac{1}{4} . \alpha ^{2} ]$ $\Rightarrow [{3 \alpha ^{2}-8 \alpha ^{3 / 2}+8=0}]$

Q43JEE Main 2020MCQ4MDifferential Equations

If a curve $\ {y}=\ {f}( {x}),$ passing through the point (1,2), is the solution of the differential equation, $2 {x}^{2} {dy}= (2 {xy}+{y}^{2}){d} {x},$ then ${f}(\frac{1}{2} )$ is equal to :

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📖 Explanation

$2 {x}^{2} {dy}= (2 {xy}+ {y}^{2} ) {dx}$ $\Rightarrow\frac{ {dy}}{ {dx}}= \frac{2 {xy}+ {y}^{2}}{2 {x}^{2}}$ $\{$ Homogeneous D.E. $\}$ $\{\text{table} \text{let} y = {xt} ; \Rightarrow \frac{ {dy}}{ {dx}}= {t}+ {x}\frac{{dt}}{ {dx} }\}$ $\Rightarrow {t}+ {x}\frac{{dt}}{\ {dx}}= \frac{2 {x}^{2} {t}+ {x}^{2} {t}^{2}}{2 {x}^{2}}$ $\Rightarrow {t}+ {x}\frac{{dt}}{{dx}}= {t}+\frac{{t}^{2}}{2}$ $\Rightarrow {x}\frac{{dt}}{ {dx}}=\frac{{t}^{2}}{2}$ $\Rightarrow 2∫\frac{{dt}}{ {t}^{2}}=∫\frac{{d} {x}}{ {x}}$ $\Rightarrow 2 (- \frac{1}{ {t}} )=l{n}( {x})+ {C}\$ $\{ \text{Put} {t}=\frac{{y}}{ {x}} \}$ $\Rightarrow -\frac{2 {x}}{ {y}}=l{n} {x}+ {C}$ $\{\text{Put} x= 1\ & \ y =2 \text{then}\text{we}\text{get} C=-1\}$ $\Rightarrow \frac{-2 {x}}{ {y}}=l{n}( {x})-1$ $\Rightarrow {y}= \frac{2 {x}}{1-l{n} {x}}$ $\Rightarrow {f}( {x})= {2 {x}}/{1-\log _{ {e}} {x}}$ so, $[ {f ( \frac{1}{2} )= \frac{1}{1+\log _{e} 2}}]$

Q44JEE Main 2020MCQ4MSequences and Series

Let $\ {S}$ be the sum of the first 9 terms of the series: $\{x+k a\}+ \{x^{2}+(k+2) a \}$ $+ \{x^{3}+(k+4) a \}+$ $\{ {x}^{4}+( {k}+6) {a} \}+.... . .$ where ${a} \ne 0$ and ${x} \ne 1 .$ If $\ {S}=\frac{{x}^{10}-{x}+45 {a}( {x}-1)}{ {x}-1},$ then ${k}$ is equal to :

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📖 Explanation

${S}=[ {x}+ {ka}+0]+\$ [{x}^{2}+ {ka}+2 {a} ]$+$[{x}^{3}+ {ka}+4 {a}]$+$[{x}^{4}+ {ka}+6 {a} ]$+.... 9 {\text{terms}}\Rightarrow {S}=({x}+ {x}^{2}+ {x}^{3}+ {x}^{4}+... . .9 {\text{terms}} )+( {ka}+ {ka}+ {ka}+ {ka}+... 9 {\text{terms}})+(0+2 {a}+4 {a}+6 {a}+..... 9 $terms$ )\Rightarrow {S}= {x}$$\frac{ {x}^{9}-1}{ {x}-1}]+9 {ka}+72 {a}\Rightarrow {S}=\frac{( {x}^{10}- {x} )+(9 {k}+72) {a}( {x}-1)}{( {x}-1)} $Compare with given \sum, then we get,$ (9 {k}+ $72)$ =45\Rightarrow [{k}=-3]$

Q45JEE Main 2020MCQ4MStraight Lines and Pair of Straight Lines

The set of all possible values of $\theta$ in the interval $(0, \pi )$ for which the points (1,2) and $(\sin \theta ,$ $\cos \theta$ ) lie on the same side of the line $x+y=$ 1 is:

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📖 Explanation

Given that both points (1,2) $\ \&\ (\sin \theta , \cos \theta )$ lie on same side of the line $x+y-1=0$ So, $(\text{Put} (1,2) { \in } { \text{given} \text{line} } )$ ${ \text{Put} }(\sin \theta , \cos \theta { \in } { \text{given} \text{line} } )>0$ $\Rightarrow(1+2-1)(\sin \theta +\cos \theta -1)>0$ $\Rightarrow \sin \theta +\cos \theta >1\{\÷\ {\text{by}} \\sqrt{2}\}$ $\Rightarrow \frac{1}{\\sqrt{2}} \sin \theta +\ \frac{1}{\\sqrt{2}} \cos \theta > \frac{1}{\\sqrt{2}}$ $\Rightarrow \sin (\theta + \frac{\pi }{4} )> \frac{1}{\\sqrt{2}}$ $\Rightarrow \frac{\pi }{4}<\theta + \frac{\pi }{4}< \frac{3 \pi }{4}$ $\Rightarrow 0 < \theta < \frac{\pi }{2}$

Q46JEE Main 2020NAT4MStatistics

If the variance of the terms in an increasing A.P., $\ {b}_{1}, \ {b}_{2}, \ {b}_{3},...{b}_{11}$ is $90,$ then the commondifference of this A.P. is _____.

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📖 Explanation

Let a be the first term and d be the common difference of the given A.P. Where $\ {d}>0$ $\overline{{X}}= {a}+$ $\frac{0+ {d}+2 {d}+...+10 {d}}{11}$ $=a+5 d$ $\Rightarrow$ varience $=\ {\sum (\overline{{X}}- {x}_{ {i}} )^{2}}{11}$ $\Rightarrow90 \\times 11= (25 {d}^{2}+16 {d}^{2}$ $+9 {d}^{2}+4 {d}^{2} ) \\times 2$ $\Rightarrow {d}=\±3\Rightarrow {d}=3$

Q47JEE Main 2020NAT4MDifferentiation

If $y=\sum _{k=1}^ {6} k \cos ^{-1}$ $\{\frac{3}{5} \cos k x-\frac{4}{5} \sin k x \},$ then $\frac{d y}{d x}$ at $x=0$ is _____.

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📖 Explanation

Put $\cos \alpha = \frac{3}{5}, \sin \alpha = \frac{4}{5}$ $\;\; 0<\alpha < \frac{\pi }{2}$ Now $\frac{3}{5} \cos {kx}- \frac{4}{5} \sin {kx}$ $=\cos \alpha \.\cos {kx}-\sin \alpha \.\sin {kx}$ $=\cos (\alpha + {kx})$ As we have to find derivate at $x=0$ We have $\cos ^{-1}(\cos (\alpha + {kx}))$ $=(\alpha + {kx})$ $\Rightarrow {y}=\sum _{k=1}^ {6} (\alpha + {kx})$ $\Rightarrow {\frac{\ {dy}}{ {dx}} |}_{ {\text{at}} {x}=0}=\sum _{k=x}^ {6}{k}$ $= \frac{6 \\times 7 \\times 13}{6}=91$

Q48JEE Main 2020NAT4MVector Algebra

Let the position vectors of points 'A' and 'B' be $\hat{i}+\hat{j}+ \hat{k}$ and $2 \hat{i}+\hat{j}+3 \hat{k},$ respectively. A point 'P' divides the line segment AB internally in the ratio $\lambda:1(\lambda>0) .$ If $\ {O}$ is the origin and ${\vec{OB}} .{\vec{OP}}-3|{\vec{OA}} \times {\vec{OP}}|^{2}=6,$ then $\lambda$ is equal to _______.

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📖 Explanation

Using section formula we get $\overline{{OP}}= \frac{2 \lambda+1}{\lambda+1} \hat{i}+ \frac{\lambda+1}{\lambda+1} \hat{j}+ \frac{3 \lambda+1}{\lambda+1} \hat{k}$ Now $\overline{{OB}}. \overline{{OP}}$ $= \frac{4 \lambda+2+\lambda+1+9 \lambda+3}{\lambda+1}$ $= \frac{14 \lambda+6}{\lambda+1}$ $\overline{{OA}} \\times \overline{{OP}}$ $=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ \frac{2 \lambda+1}{\lambda+1} & 1 & \frac{3\lambda+1}{\lambda+1}\end{vmatrix}$ $= \frac{2 \lambda+1}{\lambda+1} \hat{i}+ \frac{-\lambda}{\lambda+1} \hat{j}$ $+ \frac{-\lambda}{\lambda+1} \hat{k}$ $|\overline{{OA}} \\times \overline{{OP}}|^{2}$ $= \frac{(2 \lambda+1)^{2}+\lambda^{2}+\lambda^{2}}{(\lambda+1)^{2}}$ $=\ \frac{6 \lambda^{2}+1}{(\lambda+1)^{2}}$ $\Rightarrow \frac{14 \lambda+6}{\lambda+1}-3 \\times \frac{(6 \lambda^{2}+1 )}{(\lambda+1)^{2}}=6$ $\Rightarrow10 \lambda^{2}-8 \lambda=0$ $\Rightarrow \lambda=0, \ \frac{8}{10}=0.8$ $\Rightarrow \lambda=0.8$

Q49JEE Main 2020NAT4MBinomial Theorem

For a positive integer $\ {n}, (1+ \frac{1}{ {x}} )^{ {n}}$ is expandedin increasing powers of $\ {x} .$ If three consecutive coefficients in this expansion are in the ratio, $2: 5: 12,$ then $\ {n}$ is equal to ______.

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📖 Explanation

${ }^{n} C_{r-1}:{ }^{n} C_{r}:{ }^{n} C_{r+1}$ $=2: 5: 12$ Now $\ \frac{{ }^{n} C_{r-1}}{{ }^{n} C_{r}}=\ \frac{2}{5}$ \Rightarrow $7 r=2 n+2$ ...(1) ${{ }^{ {n}} {C}_{ {r}}}/{{ }^{ {n}} {C}_{ {r}+1}}= \frac{5}{12}$ $\Rightarrow17 \ {r}=5 \ {n}-12$ ....(2) On solving(1) &(2) $\Rightarrow {n}=118$

Q50JEE Main 2020NAT4MDefinite Integrals

Let [t] denote the greatest integer less than or equal to t. Then the value of $\int _{1} ^{2}{|2 \ {x}-[3 \ {x}]| } {dx}$ is ______.

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📖 Explanation

$3 < 3 x < 6$ Take cases when $3 < 3 \ {x} < 4,4 < 3 \ {x} < 5$ , $5 < 3 x < 6$ Now $\int _{1} ^{2}{|2 \ {x}-[3 \ {x}]| } {dx}$ $=\int _{1}^{4 / 3}(3-2 x) \ {d} x+\int _{4 / 3}^{5 / 3}(4-2 x) \ {d} x$ $+\int _{5 / 3}^{2}(5-2 x) \ {d} x$ $=\ \frac{2}{9}+ \frac{3}{9}+ \frac{4}{9}=1$

PHYSICS25 questions

Q51JEE Main 2020MCQ4MMagnetic Effects of Current and Magnetism

The figure shows a region of length ' $l$ with a uniform magnetic field of $0.3 {T}$ in it and a proton entering the region with velocity $4 \\times 10^{5} \ {ms}^{-1}$ making an angle $60°$ with the field. If the proton completes 10 revolution by thetime it cross the region shown, ' $l'$ is close to (mass of proton $=1.67 \\times 10^{-27} \ {kg},$ charge of the proton $=1.6 \\times 10^{-19} \ {C}\ )$

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📖 Explanation

${T}= \frac{2 \pi {m}}{ {qB}}$ total time ${t}=10 {T}$ Kinematics $l=\frac{{V}}{2} {t}$ $l=\frac{{V}}{2} 10 \\times \frac{2 \pi {m}}{ {qB}}$ $=4 \\times 10^{5} \\times 10 \$ $\times \frac{3.14 \\times 1.67 \\times 10^{-27}}{1.6 \\times 10^{-19} \\times 0.3}$ $=0.439$

Q52JEE Main 2020MCQ4MOscillations

The displacement time graph of a particle executing S.H.M. is given in figure : (sketch is schematic and not to scale)Which of the following statements is/are true for this motion? (A) The force is zero $t= \frac{3 {T}}{4}$ (B) The acceleration is maximum at ${t}= {T}$ (C) The speed is maximum at ${t}=\frac{{T}}{4}$ (D) The P.E. is equal to K.E. of the oscillation at $t= \frac{T}{2}$

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📖 Explanation

${ (A) } {F}= {ma}\;\;\;\ {a}=-\omega ^{2} {x}$ at $\frac{3 {T}}{4}$ displacement zero $( {x}=0),$ so ${a}=0$ ${F}=0$ (B) at ${t}={T} \;\;\;$ displacement $( {x})= {A}$ x maximum, So acceleration is maximum. (C) ${V}=\omega \sqrt{ {A}^{2}- {x}^{2}}$ ${V}_{\max }$ at ${x}=0$ ${V}_{\max }={A} \omega$ at ${t}=\frac{{T}}{4}, \ {x}=0,$ So ${V}_{ {\max}}$ (D) ${KE}={PE}$ $\therefore$ at $x=\ \frac{A}{\\sqrt{2}}$ at $t=\ \frac{T}{2}\;\; x=-A \;\;$ (So not possible)

Q53JEE Main 2020MCQ4MMagnetic Effects of Current and Magnetism

A wire carrying current I is bent in the shapeABCDEFA as shown, where rectangle ABCDAand ADEFA are perpendicular to each other. If the sides of the rectangles are of lengths aand $\ {b},$ then the magnitude and direction ofmagnetic moment of the loop ABCDEFA is :

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📖 Explanation

${M}= {NIA}$ ${N}=1$ For ${ABCD}$ ${\vec{M}}_{1}= {abI} \hat{K}$ For DEFA ${\vec{M}}_{2}= {abI} \hat{j}$ ${\vec{M}} ={\vec{M}}_{1}+{\vec{M}}_{2}$ $=\ {ab} {I}(\hat{k}+\hat{j})$ = abI, $\sqrt{2}(\frac{\hat{j}}{\sqrt{2}}+ \frac{\hat{k}}{\\sqrt{2}} )$

Q54JEE Main 2020MCQ4MCurrent Electricity

A potentiometer wire ${PQ}$ of $1 {m}$ length is connected to a standard cell ${E}_{1} .$ Another cell ${E}_{2}$ of emf $1.02 {V}$ is connected with a resistance 'r' and switch $S$ (as shown in figure). With switch ${S}$ open, the null position isobtained at a distance of $49 {cm}$ from $\ {Q}$ . The potential gradient in the potentiometer wire is :

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📖 Explanation

Balancing length is measured from $\ {P}$ . So $100-49=51 {cm}$ $\ {E}_{2}=\phi \times 51$ Where $\phi =$ Potential gradient $1.02=\phi \times 51$ $\phi =0.02$ V/cm

Q55JEE Main 2020MCQ4MThermodynamics

A heat engine is involved with exchange ofheat of $1915 {J},-40 {J},+125 {J}$ and ${QJ},$ duringone cycle achieving an efficiency of $50.0 \%$ . The value of ${Q}$ is:

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📖 Explanation

$\eta = {\text{Work done }}{\text{ Heat supplied}}$ $\frac{1}{2}=\eta = \frac{1915-40+125-{Q}}{1915+125}$ $\frac{1}{2}= \frac{2000- {Q}}{2040}$ $2040=4000-2 {Q}$ $2 \ {Q}=1960$ ${Q}=980 {J}$

Q56JEE Main 2020MCQ4MWave Optics

In a Young's double slit experiment, 16 fringesare observed in a certain segment of the screenwhen light of wavelength $700 \ {nm}$ is used. Ifthe wavelength of light is changed to $400 \ {nm}$ ,the number of fringes observed in the samesegment of the screen would be :

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📖 Explanation

Let the length of segment is " $l "$ Let $\ {N}$ is the no. of fringes in " $l"$ and $\ {w}$ is fringe width. $\to$ We can write ${N} {w}=l$ ${N}(\frac{\lambda {D}}{ {d}}\ )=l$ $\frac{{N}_{1} \lambda_{1} {D}}{ {d}}=l$ $\frac{{N}_{2} \lambda_{2} {D}}{ {d}}=l$ ${N}_{1} \lambda_{1}= {N}_{2} \lambda_{2}$ $16 \\times 700= {N}_{2} \\times 400$ ${N}_{2}=28$

Q57JEE Main 2020MCQ4MAtoms and Nuclei

In a hydrogen atom the electron makes atransition from $( {n}+1)^{ {\text{th}}}$ level to the ${n}^{ {\text{th} }}$ level. If $\ {n} > >1,$ the frequency of radiation emitted is proportional to :

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📖 Explanation

In hydrogen atom, ${E}_{ {n}}= \frac{- {E}_{0}}{ {n}^{2}}$ Where $\ {E}_{0}$ is Ionisation Energy of $\ {H}$ . $\to$ For transition from $(\ {n}+1)$ to $\ {n},$ the energyof emitted radiation is equal to the difference in energies of levels. $\Delta{E}= {E}_{ {n}+1}- {E}_{ {n}}$ $\Delta{E}= {E}_{0}( \frac{1}{ {n}^{2}}- \frac{1}{( {n}+1)^{2}} )$ $\Delta{E}= {h} v= {E}_{0}(\frac{( {n}+1)^{2}- {n}^{2}}{ {n}^{2}( {n}+1)^{2}} )$ $h v=E_{0} [{2 n+1}{n^{4} (1+ \frac{1}{n} )^{2}} ]$ $h v=E_{0} [ \frac{n (2+ \frac{1}{n} )}{n^{4} (1+ \frac{1}{n} )^{2}}]$ since ${n} > > > 1$ Hence, $\frac{1}{ {n}} \simeq 0$ ${h} v= {E}_{0}$$\frac{2}{ {n}^{3}} ]$ $v\ \alpha \ \frac{1}{\ {n}^{3}}$

Q58JEE Main 2020MCQ4MSemiconductor Electronics

In the following digital circuit, what will be the output at 'Z', when the input (A, B) are (1,0), (0,0),(1,1),(0,1):

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📖 Explanation

${Z}=(\overline{{P}+ {R}})$ ${Z}=(\overline{{P}+ {PQ}})$ ${Z}=(\overline{{P}(1+ {Q})})$ $\ {Z}=(\overline{{P}})[\ {\text{Using}}$ Identity $(1+ {A})=1]$ A B Z 1 0 0 0 0 0 1 1 1 0 1 0 ${Z}=\ov {(\ov { {AB}})}$ $\ {Z}= {AB}$ Truth table for $Z= {AB}$

Q59JEE Main 2020MCQ4MUnits and Measurements

If momentum $(\ {P}),$ area $(\ {A})$ and time $(\ {T})$ aretaken to be the fundamental quantities then thedimensional formula for energy is :

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📖 Explanation

Let $[ {E}]=[ {P}]^{ {x}}[ {A}] {y}[ {T}]^{ {z}}$ ${ML}^{2} {T}^{-2}=\$ [{MLT}^{-1} ]^{ {x}} $[ {L}^{2} ]$ {y}[ {T}]^{ {z}}{ML}^{2} {T}^{-2}= {M}^{ {x}} {L}^{ {x}+2 {y}} {T}^{- {x}+ {z}}→{x}=1→{x}+2 {y}=21+2 y=2y= \frac{1}{2}→ -x+z=-2-1+z=-2z=-1[ {E}]=[{PA}^{1 / 2} {T}^{-1} ]$

Q60JEE Main 2020MCQ4MProperties of Matter

A capillary tube made of glass of radius $0.15 \ {mm}$ is dipped vertically in a beaker filledwith methylene iodide (surfacetension $=0.05 \ {Nm}^{-1},$ density $=667 \ {kg} \ {m}^{-3} )$ which rises to height h in the tube. It is observedthat the two tangents drawn from liquid-glassinterfaces (from opp. sides of the capillary)make an angle of $60°$ with one another. Then $\ {h}$ is close to $(\ {g}=10 \ {ms}^{-2} )$

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📖 Explanation

$\ {r} \to$ radius of capillary $\ {R} \to$ Radius of meniscus. From figure, $\frac{{r}}{ {R}}=\cos 30°$ ${R}= \frac{2 {r}}{\\sqrt{3}}= \frac{2 \\times 0.15 \\times 10^{-3}}{\\sqrt{3}}$ $= \frac{0.3}{\\sqrt{3}} \\times 10^{-3} \ {m}$ Height of capillary ${h}= \frac{2 {T}}{\rho {gR}}=2 \\sqrt{3} {T}$ $h= {2 \\times 0.05}/{667 \\times 10 \times ( \frac{0.3 \\times 10^{-3}}{\\sqrt{3}} )}$ ${h}=0.087 \ {m}$

Q61JEE Main 2020MCQ4MGravitation

The height 'h' at which the weight of a bodywill be the same as that at the same depth 'h'from the surface of the earth is (Radius of theearth is $\ {R}$ and effect of the rotation of the earthis neglected):

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📖 Explanation

$◆ {M}=$ mass of earth $\ {M}_{1}=$ mass of shaded portion $\ {R}=$ Radius of earth $\ ◆{M}_{1}=\frac{{M}}{ \frac{4}{3} \pi {R}^{3}} .\frac{4}{3} \pi ( {R}- {h})^{3}$ $= \frac{M(R-h)^{3}}{R}$ ◆ Weight of body is same at ${P}$ and ${Q}$ i.e. ${mg}_{ {P}}= {mg}_{ {Q}}$ ${g}_{ {P}}= {g}_{ {Q}}$ $\frac{{GM}_{1}}{( {R}- {h})^{2}}=\frac{{GM}}{( {R}+ {h})^{2}}$ $\frac{{GM}( {R}- {h})^{3}}{( {R}- {h})^{2} {R}^{3}}=\frac{{GM}}{( {R}+ {h})^{2}}$ $( {R}-{h})( {R}+ {h})^{2}= {R}^{3}$ ${R}^{3}- {hR}^{2}- {h}^{2} {R}- {h}^{3}$ $+2 {R}^{2} {h}-2 {Rh}^{2}= {R}^{3}$ ${R}^{2}-{Rh}^{2}- {h}^{3}=0$ ${R}^{2}-{Rh}- {h}^{2}=0$ $h^{2}+R h-R^{2}=0$ $\Rightarrow h= {-R ±\sqrt{R^{2}+4 R^{2}}}/{2}$ ie $h= \frac{-R+\\sqrt{5} R}{2}$ $=(\frac{\\sqrt{5}-1}{2} ) R$

Q62JEE Main 2020MCQ4MProperties of Matter

An ideal gas in a closed container is slowly heated. As its temperature increases, which of the following statements are true? (A) the mean free path of the molecules decreases. (B) the mean collision time between the molecules decreases. (C) the mean free path remains unchanged. (D) the mean collision time remainsunchanged.

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📖 Explanation

The mean free path of molecules of an ideal gas is given as: $\lambda=\frac{{V}}{\\sqrt{2} \pi {d}^{2} {N}}$ where : $\ {V}=$ Volume of container $\ {N}=$ No of molecules Hence with increasing temp since volume of container does not change (closed container), so mean free path is unchanged. Average collision time $= {\text{ mean free path }}/{ {V}_{ {av}}}= \frac{\lambda}{(\text{avg speed of molecules})}$ $\because$ avg speed $\alpha \sqrt{ {T}}$ $\therefore$ Avg coll. time $\alpha {1}/{\\sqrt{ {T}}}$ Hence with increase in temperature the average collision time decreases.

Q63JEE Main 2020MCQ4MElectrostatic Potential and Capacitance

A charge $Q$ is distributed over two concentricconducting thin spherical shells radii $\ {r}$ and $\ {R}$ $(\ {R} > \ {r}) .$ If the surface charge densities on thetwo shells are equal, the electric potential at thecommon centre is :

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📖 Explanation

Let the charges on inner and outer spheres are $\ {Q}_{1}$ and $\ {Q}_{2}$ since charge density '\sigma ' is same for bothspheres, so $\sigma =\frac{{Q}_{1}}{4 \pi {r}^{2}}=\frac{{Q}_{2}}{4 \pi {R}^{2}}$ $\Rightarrow \frac{ {Q}_{1}}{ {Q}_{2}}=\frac{{r}^{2}}{ {R}^{2}}$ ${Q}_{1}+ {Q}_{2}= {Q} \Rightarrow \frac{{Q}_{2} {r}^{2}}{ {R}^{2}}+ {Q}_{2}= {Q}$ $\Rightarrow {Q}_{2}=\frac{{QR}^{2}}{({r}^{2}+ {R}^{2} )}$ ${Q}_{1}=\frac{{r}^{2}}{ {R}^{2}} .\frac{{QR}^{2}}{({R}^{2}+ {r}^{2} )}$ $=\frac{{Qr}^{2}}{({R}^{2}+{r}^{2} )}$ Potential at centre 'O' $=\frac{{kQ}_{1}}{ {r}}+\frac{{kQ}_{2}}{ {R}}$ $= {k} [\frac{{Qr}^{2}}{ {r}({R}^{2}+ {r}^{2} )}+\frac{{QR}^{2}}{ {R}({R}^{2}+ {r}^{2} )} ]$ $=\frac{{kQ}( {r}+ {R})}{({R}^{2}+ {r}^{2} )}$ $= \frac{1}{4 \pi \in _{0}} \frac{( {R}+ {r})}{({R}^{2}+ {r}^{2} )} {Q}$

Q64JEE Main 2020MCQ4MAlternating Current

An inductance coil has a reactance of $100 \Omega$ . When an AC signal of frequency $1000 \ {Hz}$ is applied to the coil, the applied voltage leads the current by $45°.$ The self-inductance of the coil is :

🚀 Solve in Practice Mode

📖 Explanation

◆ Reactance of inductance coil $=\sqrt{ {R}^{2}+ {x}_{ {L}}^{2}}=100$ ......(i) ◆ ${f}=1000 {Hz}$ of applied ${AC}$ signal ◆ Voltage leads current by $45°$ $\tan 45°={{iX}_{ {L}}}/{ {iR}}= \frac{\omega {L}}{ {R}}$ ie $R=X_{L}=\omega L$ Putting in eqn (i) $: \sqrt{ {X}_{ {L}}^{2}+ {X}_{ {L}}^{2}}=100$ $\\sqrt{2} {X}_{ {L}}=100 \Rightarrow {X}_{ {L}}=50 \\sqrt{2}$ ie $\omega {L}=50 \\sqrt{2}$ ${L}= \frac{50 \\sqrt{2}}{\omega }= \frac{50 \\sqrt{2}}{2 \pi {f}}$ $= \frac{25 \sqrt{2}}{\pi \times 1000} \ {H}$ $=1.125 \\times 10^{-2} \ {H}$

Q65JEE Main 2020MCQ4MRotational Motion

Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres. The moment of inertia and angularvelocity of the first disc are $0.1 \ {kg}-\ {m}^{2}$ and 10 rad $\ {s}^{-1}$ respectively while those for the second one are $0.2 \ {kg}-\ {m}^{2}$ and $5 \ {\text{rad}} \ {s}^{-1}$ respectively. At some instant they get stuck together and start rotating as a single system about their common axis with some angular speed. The Kinetic energy of the combined system is :

🚀 Solve in Practice Mode

📖 Explanation

◆ Both discs are rotating in same sense ◆ Angular momentum conserved for the system i.e. ${L}_{1}+ {L}_{2}= {L}_{\text{final}}$ ${I}_{1} \omega _{1}+ {I}_{2} \omega _{2}= ( {I}_{1}+ {I}_{2} ) \omega _{ {f}}$ $0.1 \\times 10+0.2 \\times 5$ $=(0.1+0.2) \\times \omega _{ {f}}$ $\omega _{ {f}}= \frac{20}{3}$ ◆ Kinetic energy of combined disc system $\Rightarrow \frac{1}{2}({I}_{1}+ {I}_{2} ) \omega _{ {f}}^{2}$ $= \frac{1}{2}(0.1+0.2) .( \frac{20}{3} )^{2}$ $= \frac{0.3}{2} \\times \frac{400}{9}= \frac{120}{18}= \frac{20}{3} \ {J}$

Q66JEE Main 2020MCQ4MMotion in a Plane

A small point mass carrying some positive charge on it, is released from the edge of a table. There is a uniform electric field in this region in the horizontal direction. Which of thefollowing options then correctly describe the trajectory of the mass ? (Curves are drawn schematically and are not to scale).

🚀 Solve in Practice Mode

Q67JEE Main 2020MCQ4MElectrostatic Potential and Capacitance

A $10 \mu{F}$ capacitor is fully charged to a potential difference of $50 \ {V}$ . After removing the source voltage it is connected to an unchargedcapacitor in parallel. Now the potential difference across them becomes $20 \ {V}$ . Thecapacitance of the second capacitor is:

🚀 Solve in Practice Mode

📖 Explanation

Initially ◆ Charge on capacitor $10 \mu{F}$ ${Q}= {CV}=(10 \mu{F})(50 {V})$ ${Q}=500 \mu{C}$ ◆ Final Charge on $10 \mu{F}$ capacitor ${Q}= {CV}=(10 \mu{F})(20 {V})$ ${Q}=200 \mu{C}$ ◆ From charge conservation, Charge on unknown capacitor ${C}=500 \mu{C}-200 \mu{C}$ $=300 \mu{C}$ $\Rightarrow$ Capacitance $( {C})=\frac{{Q}}{ {V}}= \frac{300 \mu{C}}{20 {V}}$ $=15 \mu{F}$

Q68JEE Main 2020MCQ4MProperties of Matter

When the temperature of a metal wire isincreased from $0°{C}$ to $10°{C},$ its length increasesby $0.02 \% .$ The percentage change in its massdensity will be closest to:

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📖 Explanation

Given $\frac{\Delta{L}}{ {L}}=0.02 \%$ $\therefore \Delta{L}= {L} \alpha \Delta{T}$ $\Rightarrow \frac{\Delta{L}}{ {L}}=\alpha \Delta{T}=0.02 \%$ $\therefore \beta =2 \alpha$ (Areal coefficient of expansion) $\Rightarrow \beta \Delta{T}=2 \alpha \Delta{T}=0.04 \%$ Volume $=$ Area $\\times$ Length Density $(\rho )= \frac{\text{Mass }}{\text{Volume}}$ $= \frac{\text{ Mass }}{\text{ Area} \\times \text{ Length }}=\frac{{M}}{ {AL}}$ $\Rightarrow \frac{\Delta \rho }{\rho }= \frac{\DeltaM}{ {M}}- \frac{\Delta{A}}{ {A}}- \frac{\Delta{L}}{ {L}}$ ( Mass remains constant $)$ $\Rightarrow ( \frac{\Delta \rho }{\rho } )= \frac{\Delta{A}}{ {A}}+ \frac{\Delta{L}}{ {L}}$ $=\beta \Delta{T}+\alpha \Delta{T}$ $=0.04 \%+0.02 \%$ $=0.06 \%$

Q69JEE Main 2020MCQ4MElectromagnetic Waves

In a plane electromagnetic wave, the directionsof electric field and magnetic field arerepresented by $\hat{k}$ and $2\hat{i}-2 \hat{j},$ respectively. What is the unit vector along direction of propagation of the wave.

🚀 Solve in Practice Mode

📖 Explanation

$\hat{E}=\hat{k}$ ${\vec{B}}=2 \hat{i}-2 \hat{j}$ $\Rightarrow \hat{B}=\frac{\vec{B}}{| {B}|}= \frac{2 \hat{i}-2 \hat{j}}{2 \\sqrt{2}}$ $\Rightarrow \hat{B}= \frac{1}{\\sqrt{2}}(\hat{i}-\hat{j})$ Direction of wave propagation $=\hat{C}=\hat{E} \times \hat{B}$ ${ \hat{C}}=\hat{k} \times [\frac{1}{\\sqrt{2}}(\hat{i}-\hat{j}) ]$ $\hat{C}= \frac{1}{\\sqrt{2}}(\hat{k} \times \hat{i}-\hat{k} \times \hat{j})$ $\hat{C}= \frac{1}{\\sqrt{2}}(\hat{i}+\hat{j})$

Q70JEE Main 2020MCQ4MDual Nature of Matter and Radiation

A particle is moving 5 times as fast as anelectron. The ratio of the de-Brogliewavelength of the particle to that of the electronis $1.878 \\times 10^{-4}$ . The mass of the particle is closeto :

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📖 Explanation

Let mass of particle $=\ {m}$ Let speed of $\ {e}^{-}=\ {V}$ $\Rightarrow$ speed of particle $=5 \ {V}$ Debroglie wavelength $\lambda_{ {d}}=\frac{{h}}{ {P}}=\frac{{h}}{ {mv}}$ $\Rightarrow (\lambda_{ {d}} )_{ {P}}= \frac{ {h}}{ {m}(5 {V})}$ ......(1) $\Rightarrow (\lambda_{ {d}} )_{ {e}}= {{h}}/{ {m}_{ {e}}.{V}}$ ......(2) According to question $\frac{(1)}{(2)}={{m}_{ {e}}}/{5 {m}}=1.878 \\times 10^{-4}$ $\Rightarrow {m}={{m}_{ {e}}}/{5 \\times 1.878 \\times 10^{-4}}$ $\Rightarrow{m}= \frac{9.1 \\times 10^{-31}}{5 \\times 1.878 \\times 10^{-4}}$ $\Rightarrow {m}=9.7 \\times 10^{-28} \ {kg}$

Q71JEE Main 2020NAT4MCurrent Electricity

An ideal cell of emf $10 \ {V}$ is connected in circuitshown in figure. Each resistance is $2 \Omega .$ Thepotential difference (in $\ {V}$ ) across the capacitorwhen it is fully charged is _____.

🚀 Solve in Practice Mode

📖 Explanation

◆ $\ {R}_{1}$ to $\ {R}_{5} \to$ each $2 \Omega$ ◆ Cap. is fully charged ◆ So no current is there in branch ADB ◆ Effective circuit of current flow : ${R}_{ {eq}}=(\frac{4 \\times 2}{4+2} )+2$ ${R}_{ {eq}}=\frac{4}{3}+2= \frac{10}{3} \Omega$ ${i}= \frac{10}{10 / 3}=3 \ {A}$ So potential different across AEB $\Rightarrow 2 \\times 1+2 \\times 3=8 \ {V}$ Hence potential difference across Capacitor $=\Delta {V}= {V}_{{AEB}}=8 {V}$

Q72JEE Main 2020NAT4MRay Optics and Optical Instruments

A light ray enters a solid glass sphere of refractive index $\mu=\\sqrt{3}$ at an angle ofincidence $60° .$ The ray is both reflected andrefracted at the farther surface of the sphere. The angle (in degrees) between the reflectedand refracted rays at this surface is _____.

🚀 Solve in Practice Mode

📖 Explanation

By Snell's law at $\ {A}$ : $1 \times \sin 60°=\\sqrt{3} \\times \sin \theta$ $\frac{\\sqrt{3}}{2}=\\sqrt{3} \sin \theta$ $\sin \theta = \frac{1}{2} \Rightarrow \theta =30°$ So at B : $\theta +60°+\alpha =180°$ $30°+60°+\alpha =180°$ $\alpha =90°$

Q73JEE Main 2020NAT4MCentre of Mass and Collision

A particle of mass $\ {m}$ is moving along the $\ {x}$ -axis with initial velocity $undefined$ . It collides elastically with a particle of mass $10 \ {m}$ at rest and then moves with half its initial kinetic energy (see figure). If $\sin \theta _{1}=\\sqrt{ {n}}\sin \theta _{2}$ then value of $\ {n}$ is _____.

🚀 Solve in Practice Mode

📖 Explanation

By momentum conservation along $y:$ $\ {m}_{1} {u}_{1} \sin \theta _{1}= {m}_{2} {u}_{2} \sin \theta _{2}$ i.e. ${\mu}_{1} \sin \theta _{1}=10 {\mu}_{2} \sin \theta _{2}$ \Rightarrow $[u_1\sin\theta _1=10u_2\sin\theta _2]$ ......(i) ${kf}_{\ {m}_{1}}= \frac{1}{2} {ki}_{ {m}_{1}}$ i.e. $\frac{1}{2} {\mu}_{1}^{2}={1}{2}\times \frac{1}{2} {\mu}^{2}$ i.e. $[u_{1}= \frac{u}{\\sqrt{2}}]$ ........(ii) Also collision is elastic $: {k}_{ {i}}= {k}_{ {f}}$ $\frac{1}{2} {\mu}^{2}= \frac{1}{2} {\mu}_{1}^{2}+ \frac{1}{2} .10 {m} .{u}_{2}^{2}$ $\frac{1}{2} {\mu}^{2}= \frac{1}{2} \times \frac{1}{2} {\mu}^{2}$ $+ \frac{1}{2} \\times 10 {m}.{u}_{2}^{2}$ $\frac{1}{4} {\mu}^{2}= \frac{1}{2} \\times 10 \\times {\mu}_{2}^{2}$ $[{u}_{2}=\frac{{u}}{\\sqrt{20}}]$ ......(iii) Putting(ii) & (iii) in (i) $\frac{ {u}}{\\sqrt{2}} \sin \theta _{1}=10.\frac{{u}}{\\sqrt{20}} \sin \theta _{2}$ $[\sin \theta _{1}=\sqrt{10} \sin \theta _{2} ] \to$ Hence $\ {n}=10$

Q74JEE Main 2020NAT4MCentre of Mass and Collision

A square shaped hole of side $l= \frac{\ {a}}{2}$ is carvedout at a distance $\ {d}=\frac{{a}}{2}$ from the centre 'O' ofa uniform circular disk of radius a. If thedistance of the centre of mass of the remainingportion from $\ {O}$ is $-\frac{{a}}{ {X}},$ value of ${X}$ (to thenearest integer) is ______.

🚀 Solve in Practice Mode

📖 Explanation

${X}_{ {\text{com}}}={{m}_{1} {x}_{1}- {m}_{2} {x}_{2}}{ {m}_{1}- {m}_{2}}$ where : ◆ $\ {m}_{1}=$ mass of complete disc ◆ ${m}_{2}=$ removed mass ◆ Let $\sigma =$ surface mass density of disc material wrt 'O' : ${X}_{ {\text{com} }}= {\sigma \pi {a}^{2}( {O})-\sigma .\frac{{a}^{2}}{4} .{d}} /{\sigma \pi {a}^{2}-\sigma \frac{{a}^{2}}{4}}$ $= {-\frac{{a}^{2}}{4} {d}}/{\pi {a}^{2}-\frac{{a}^{2}}{4}}$ $=\frac{- {d}}{4 \pi -1}=-\frac{{a}}{2(4 \pi -1)}$ So, $X=2(4 \pi -1)$ $=(8 \pi -2)=23.12$ So, nearest integer value of $X=23$

Q75JEE Main 2020NAT4MWaves

A wire of density $9 \\times 10^{-3} \ {kg} \ {cm}^{-3}$ is stretched between two clamps $1 \ {m}$ apart. The resulting strain in the wire is $4.9 \\times 10^{-4}$ . The lowest frequency of the transverse vibrations in the wire is (Young's modulus of wire $Y=9\times 10 ^10Nm^{-2}),$ (to the nearest integer), ______.

🚀 Solve in Practice Mode

📖 Explanation

$\rho _{ {\text{wire} }}=9 \\times 10^{-3} \frac{{kg}}{\ {cm}^{3}}$ $= \frac{9 \\times 10^{-3}}{10^{-6}} \ {kg} / {m}^{3}$ $=9000 {kg} / {m}^{2}$ (A = CSA of wire) (Y = $9 \\times 10^{10} \ {Nm}^{2}$ ) (Strain $=4.9 \\times 10^{-4}$ ) $\Rightarrow {L}=1 \ {m}= \frac{\lambda}{2} \Rightarrow \lambda=2 \ {m}$ $\Rightarrow {v}=\ {f} \lambda \Rightarrow \\sqrt{\frac{{T}}{\mu}}=\ {f} \lambda$ Where $Y=\frac{{T} / {A}}{\text{strain }} \Rightarrow {T}= {Y} .$ A. strain $\Rightarrow \sqrt{ \frac{\text{ Y.A. strain}}{ {m} / {L}}}= {f} \\times 2$ $\Rightarrow \sqrt{ \frac{\text{ Y.A.L. strain }}{{M}}}= {f} \\times 2$ $\Rightarrow \sqrt{\frac{{Y} \times {V}\times \text{ strain }}{ {M}}}={f}\times 2$ $\Rightarrow \sqrt{\frac{{Y} \times \text{ strain }}{\rho }}= {f} \\times 2$ ${f}= \frac{1}{2}.\sqrt{{{Y}\times {\text{strain}}}/{\rho }}$ $= \frac{1}{2} \\sqrt{ \frac{9 \\times 10^{10} \\times 4.9 \\times 10^{-4}}{9000}}$ ${f}= \frac{1}{2}.\sqrt{ \frac{9 \\times 10^{3}}{9} \\times 4.9}$ $=\ \frac{1}{2} \\sqrt{4900}=\ \frac{70}{2}=35 \ {Hz}$

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