1 Sep 2021 Shift 2

Given, the side of the cube, $\text{s}=0.5 {\text{m}}$ Electric field, $\text{E}=150 \text{y}^{2} \hat{j}$ The direction of electric field is as shown in the below figure, At bottom surface, $\text{y}=0$ As we know that, the expression of electric flux, $\phi=\text{E} .\text{A} \cos \theta$ Here, E is the electric field passing through the cube and A is the surface area of the cube. Substituting the values in the above equations, we get $\phi=150 \text{y}^{2} .(0.5 \times 0.5) \times \cos180°$ $=150(0)^{2} .(0.25) \times (-1)=0$ Hence, the electric flux is zero at the bottom surface. At the top surface, $\text{y}=0.5 {\text{m}}$ Electric field, $\text{E}=150 {\text{y}}^{2} \hat{j}=150(0.5)^{2}=37.5 {\text{N}} / {\text{C}}$ Electric flux at the top surface, $\phi=\text{E} .\text{A} \cos \theta$ $=(37.5) .(0.5 \times 0.5) \cos 0°$ $=9.375 {\text{N}} / {\text{C}}-{\text{m}}^{2}$ By using the Gauss's law, $\phi={\text{Q}_{{\in }}}/{{\epsilon_0}}$ Here, $\text{Q}_{{\in }}=$ net charge enclosed in the cube and ${\epsilon_0}=$ permittivity of the free space. Substituting the values in the above equation, we get $9.375={\text{Q}_{{\in }}}/{8.85 \times 10^{-12}}$ $\text{Q}_{{\in }}=8.3 \times 10^{-11} {\text{C}}$ The charge inside the cube is $8.3 \times 10^{-11} {\text{C}}$.

According to given circuit diagram, equivalent resistance between point P and Q. $\text{R}_{\text{P}\text{Q}}=(4+4) \||(4+4)$ $=\frac{8 \times 8}{8+8}=4 \Omega$ The equivalent circuit can be drawn as, Equivalent resistance, $\text{R}_{{\text{eq}}}=4+1=5 \Omega$ Magnetic field, B = 5T The side of the square loop, l = 20 cm = 0.20 m The steady value of the current, $\text{I}=2 {\text{mA}}=2 \times 10^{-3} {\text{A}}$ Induced emf, $\text{e}=\text{B} \text{v}_{0} \text{l}$ Induced current, $\text{I}={{\text{e}}}/{\text{R}_{{\text{eq}}}}$ Substituting the values in the above equation, we get $2 \times 10^{-3}=\frac{5 \times v_{0} \times 0.2}{5}$ \Rightarrow $\text{v}_{0}=10^{-2} {\text{m}} / {\text{s}}=1 {\text{cm}} / {\text{s}}$ ∴ The value of $v_{0}=1 {\text{cm}} /$ so that a steady current of $2 {\text{mA}}$ flows in the loop.

Given, Planck's constant, $\text{h}=6.6 \times 10^{-34} {\text{Js}}$ Boltzmann constant, $\text{k}_{\text{B}}=1.38 \times 10^{-23} {\text{J}} / {\text{K}}$ Mass of an electron, $\text{m}_{\text{e}}=9 \times 10^{-31} {\text{kg}}$ Temperature of an ideal gas, $\text{T}=300 {\text{K}}$ As we know that, de-Broglie wavelength, $\lambda=\frac{\text{h}}{\text{m} \text{v}}={\text{h}}/{\sqrt{2 \text{mE}}}$ ... (i) Here, E is the kinetic energy. $\text{E}={3 \text{K}_{\text{B}} \text{T}}/{2}$ Substituting value of $\text{E}$ in Eq. (i), we get, $\lambda={\text{h}}/{\sqrt{3 \text{mK}_{\text{B}} \text{T}}}$ Substituting the given values in the above equation, we get $\lambda={6.6 \times 10^{-34}}/{\sqrt{3 \times 9 \times 10^{-31} \times 138 \times 10^{-23} \times 300}}$ = 6.26 nm ∴ The corresponding de-Broglie wavelength of an electronapproximately at 300 K is 6.26 nm.

Given, the mass of the body = m The height from which the body dropped = h The speed of the body when reached the ground, $\text{v}_{\text{f}}=0.8 \sqrt{\text{gh}}$ Initial velocity of the body, v = 0 m / s Using the work-energy theorem, Work done by gravity + Work done by air-friction = Final kinetic energy - Initial kinetic energy $\text{W}_{{\text{mg} }}+\text{W}_{{\text{air}- \text{friction} }}=\frac{1}{2} \text{m} \text{v}_{\text{f}}^{2}-\frac{1}{2} \text{m} \text{v}_{\text{i}}^{2}$ Here, work done by gravity = mgh \Rightarrow $\text{m} \text{g} \text{h}+\text{W}_{{\text{air}-\text{friction} }}=\frac{1}{2} \text{m}(0.8 \sqrt{\text{g} \text{h}})^{2}-\frac{1}{2} \text{m}(0)^{2}$ \Rightarrow $\text{W}_{{\text{air} - \text{fricion} }}=\frac{0.64 \text{m} \text{g} \text{h}}{2}-\text{m} \text{g} \text{h}$ = 0.32 mgh - mgh = -0.68 mgh The value of the work done by the air friction is -0.68 mgh.

Here, two projectiles are projected at angles 42° and 48° withsame initial velocity. As we know the expression of range of projectile, Range $=\frac{\text{u}^{2} \sin 2 \theta}{\text{g}}$ At $\theta_1$ = 42° Range, $\text{R}_{1}=\frac{\text{u}^{2} \sin 2(42)°}{\text{g}}=\frac{0.99 \text{u}^{2}}{\text{g}}$ At $\theta_2$ = 48° Range, $\text{R}_{2}=\frac{\text{u}^{2} \sin 2(48)°}{\text{g}}=\frac{0.99 \text{u}^{2}}{\text{g}}$ The range of the projectile is same for the two projectiles. Therefore, $\text{R}_1 = \text{R}_2$ Now, as we know the expression of height of the projectile, $\text{H}_{\max }=\frac{\text{u}^{2} \sin \theta}{2 \text{g}}$ At 42°, $\text{H}_{\max }=\text{H}_{1}=\frac{\text{u}^{2} \sin 42°}{2 \text{g}}=\frac{0.669 \text{u}^{2}}{2 \text{g}}$ At 48°, $\text{H}_{\max }=\text{H}_{2}=\frac{\text{u}^{2} \sin 48°}{2 \text{g}}=\frac{0.743 \text{u}^{2}}{2 \text{g}}$ Higher the value of q higher the value of maximum height.Therefore, $\text{H}_1 < \text{H}_2$.

Here, both block and wedge are moving. Consider the acceleration of the block with respect to the wedge is $\text{a}_1$ and the acceleration of the wedge is $\text{a}_2$. Given, mass of the wedge, M = 16 kg and mass of block, m = 8 kg Let's draw the free body diagram of the wedge, In the x-directions, N cos60º = $\text{Ma}_2$ $\text{N}(0.5)=16 \times \text{a}_{2}$ \Rightarrow $\text{N}=32 \text{a}_{2}$ Now, draw the free body diagram of the block with respect to thewedge. Along the perpendicular to the inclined plane, $\text{N} = 8\text{g} \cos30° - 8\text{a}2\sin30º$ $32 \text{a}_{2}=4 \sqrt{3} \text{g}-4 \text{a}_{2}$ $36 \text{a}_{2}=4 \sqrt{3} \text{g}$ \Rightarrow $\text{a}_{2}=\frac{\sqrt{3}}{9} \text{g}$ Along the inclined plane, $\text{m} \text{g} \sin 30° +\text{m}\text{a}_{2} \cos 30° =\text{m} \text{a}_{1}$ $8 \text{g} \times \frac{1}{2}+8 \times \frac{\sqrt{3}}{9} \text{g} \times \frac{\sqrt{3}}{2}=8 \text{a}_{1}$ \Rightarrow $\text{a}_{1}=\frac{2 \text{g}}{3}$ ∴ The acceleration of the block with respect to the wedge is ${2\text{g}}/ {3}$.

Given, the length of the water pipe, L = 1 m The cross-sectional area of the water pipe, A$=1 {\text{cm}}^{2}=10^{-4} {\text{m}}^{2}$ The temperature of the ice = -10ºC Current passing in the conductor, I = 0.5 A Resistance of the conductor, R = 4 kΩ The latent heat of fusion for ice, $,\text{L}_{\text{i}}=3.33 \times 10^{5} {\text{J}} / {\text{kg}}$ The density of the ice, d $=1000 {\text{kg}} / {\text{m}}^{3}$ The specific heat of the ice, $\text{c}_{\text{p}, { \text{ice} }}=2 \times 10^{3} {\text{J}} / {\text{kg}}$ Heat required to melt the ice at 10°C to 0°C $\text{Q}=\text{m} \text{c}_{\text{p}} \Delta \text{T}+\text{m} \text{L}_{\text{i}}$ $\text{Q}=\text{d} \text{V} \text{c}_{\text{p}} \Delta \text{T}+\text{d} \text{V} \text{L}_{\text{f}}$ $=1000 \times 10^{-4} \times 2 \times 10^{3} \times (10) +1000 \times 10^{-4} \times 3.33 \times 10^{5}$ $(\because \text{V}=\text{A} \times \text{L})$ = 35300 J According to the Joule's law of heating, $\text{H}=\text{I}^{2} \text{R} \text{t}$ \Rightarrow $35300=(0.5)^{2}(4000)(\text{t})$ t = 35.3 s Thus, the minimum time required to melt the ice is 35.3 s.

The given formula of Young's modulus of elasticity, $\text{Y}=\frac{\text{m} \text{g} \text{L}^{3}}{4 \text{b} \text{d}^{3} \delta}$ where, Y = Young's modulus of elasticity, m = mass of the bar, L = length of the bar b = breadth of the bar d = thickness of the bar and $\delta$ = depression of the bar. There is no error in the value of the g. The fractional error in the measurement of Y, $\frac{\Delta\text{Y}}{\text{Y}}$ $=\frac{\Delta\text{M}}{\text{M}}+3 \frac{\Delta\text{L}}{\text{L}}+\frac{\Delta\text{b}}{\text{b}}+3 \frac{\Delta\text{d}}{\text{d}}+\frac{\Delta\delta}{\delta}$ Substituting the values in the above expression, we get $\frac{\Delta\text{Y}}{\text{Y}}=\frac{10^{-3}}{2}+3 \frac{(1 \times 10^{-3})}{1}+\frac{0.1 \times 10^{-3}}{4 \times 10^{-2}} +3 \frac{(0.01 \times 10^{-3})}{0.4 \times 10^{-2}}+\frac{(0.01 \times 10^{-3})}{5 \times 10^{-3}}$ $\frac{\Delta\text{Y}}{\text{Y}}=0.0155$ The fractional error in the measurement of the Young's modulus is 0.0155.

Given, $\text{R}_{1}=(4 \\pm 0.8) \Omega$ $\text{R}_{2}=(4 \\pm 0.4) \Omega$ Equivalent resistance when the resistors are connected in parallel isgiven by ${1}/{\text{R}_{{\text{eq}}}}=\frac{1}{\text{R}_{1}}+\frac{1}{\text{R}_{2}}$ \Rightarrow ${1}/{\text{R}_{\text{eq}}}=\frac{1}{4}+\frac{1}{4}$ $\text{R}_{{\text{eq}}}=2 \Omega$ Now, ${\Delta \text{R}_{{\text{eq}}}}/{\text{R}_{{\text{eq}}}^{2}}$ $=\frac{\Delta \text{R}_{1}}{\text{R}_{1}^{2}}+\frac{\Delta \text{R}_{2}}{\text{R}_{2}^{2}}$ Substituting the values in the above equation, we get ${\Delta \text{R}_{{\text{eq}}}}/{4}=\frac{0.8}{16}+\frac{0.4}{16}$ $\Delta \text{R}_{\text{eq}}=0.3 \Omega$ ∴ The equivalent resistance in parallel combination is $\text{R}_{{\text{eq}}}=(2 \\pm 0.3) \Omega$.

Given, $(\tau _{1 / 2})_{x}=(\tau )_{y}$...(i) Here, $\tau _{1/2}$ = half-life period of radioactive elementand t = mean life period of radioactive element. As we know the expression, Half-life of the radioactive element x, $\tau _{1 / 2}=\frac{\ln (2)}{\lambda_{x}}$ Mean life of the radioactive element y, $\tau =\frac{1}{\lambda_{y}}$ Substituting the values in Eq (i), we get $\frac{\ln (2)}{\lambda_{x}}=\frac{1}{\lambda_{y}}$ $\lambda_{x}=0.693 \lambda_{y}$ Initially they have same number of atoms, $\text{N}_{x}=\text{N}_{y}=\text{N}_{0}$ As we know that, Activity, A = lambda N As, $\lambda_{x} < \lambda_{y}$ \Rightarrow $\text{A}_{x} < \text{A}_{y}$ Therefore, y will decay faster than x.

The relation between magnetisation and magnetising field, $\text{M}=\chi \text{H}$ For diamagnetic material, the magnetic susceptibility $(\chi )$ is negative. The correct graph is (A) and (C)

Let us draw the diagram of glass tumbler. Consider the actual height of the tumbler be H. The refractive index of the water, $\\mu = 4 / 3$ The refractive index of the air, $\\mu =1$ As we know that, $\\mu _{{\text{water} }}={\text{H}_{{\text{real} }}}/{\text{H}_{{\text{appearent} }}}$ $\frac{4}{3}={\text{H}}/{\text{H}_{{\text{appearent} }}}$ \Rightarrow $\text{H}_{{\text{appearent} }}=\frac{3 \text{H}}{4}$ The height of air observed by observer = 17.5 -H According to the above figure, $\frac{3 \text{H}}{4}=17.5-\text{H}$ ⇒ H = 10.11 cm ≃ 10 cm The actual height of the water in tumbler is 10 cm.

The diode $\text{D}_{1}$ and $\text{D}_{2}$ are forward bias and $\text{D}_{3}$ diode is reverse bias. For forward bias, R = 30 Ω For reverse bias, R = infinite Now, current $\text{I}_{1}$ is drawn from the $200 {\text{V}}$ battery as shown in the figure. Apply the KVL rule, $-(\frac{\text{I}_{1}}{2}) 30-(\frac{\text{I}_{1}}{2}) 130-20 \text{V}_{1}+200$ $=0$ \Rightarrow $\text{I}_{1}=2 {\text{A}}$

In the steady state, the inductor behaves as a conducting wire.So, the equivalent circuit becomes as shown below. Equivalent resistance of $\text{R}_{1}, \text{R}_{2}$ and $\text{R}_{3}$ is given as, $\frac{1}{\text{R}'}=\frac{1}{\text{R}_{1}}+\frac{1}{\text{R}_{2}}+\frac{1}{\text{R}_{3}}$ $\frac{1}{\text{R}'}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}$ \Rightarrow $\text{R}_{{\text{eq} }}=1 \Omega$ Now, the circuit becomes $\text{R}_{\text{eq}}=2+1=3 \Omega$ Voltage, V = 30 V Using the Ohm's law, V = IR 30 = I (3) ⇒ I = 10 A ∴ The steady state current is 10 A.

Let's draw the free body diagram, (During the upward direction) F = 2N = (+ ve) constant (During the downward direction) F = 2N = (- ve) constan During the upward motion, the force is positive constant and duringthe downward motion the force is negative constant. Hence, the correct graph is (b).

From the potential energy curve, $\text{U}_{\max }=\frac{1}{2} \text{k} \text{A}^{2}$ $10=\frac{1}{2} \text{k}(2)^{2}$ \Rightarrow k = 5 N / m The length of the simple pendulum, L = 4 m Time period of spring, $\text{T}=2 \pi\sqrt{\frac{k}{m}}$ Time period of simple pendulum, $\text{T}=2 \pi\sqrt{\frac{\text{I}}{g}}$ The time period of simple pendulum is same as the time period ofthe spring oscillation. $\Rightarrow 2 \pi\sqrt{\frac{\text{I}}{g}}=2 \pi\sqrt{\frac{k}{m}}$ Substituting the values in the above equations, we get $2 \pi\sqrt{\frac{4}{g}}=2 \pi\sqrt{\frac{5}{5}}$ $g=4 {\text{m}} / {\text{s}}^{2}$ ∴ The acceleration due to the gravity on the planet is $4 {\text{m}}/ {\text{s}}^{2}$.

Given, the peak voltage of the battery, $\text{V}_0$ = 20V The voltage of the battery, V = 2V Time, $\text{t}=1 \\mu {\text{s}}=1 \times 10^{-6} {\text{s}}$ Resistance of the capacitor, R = 10 Ω As we know that, $\text{V}=\text{V}_{0}(1-{\text{e}}^{-\text{t} /\text{R} \text{C}})$ Substituting the values in the above equation, we get $2=20(1-{e}^{-\text{t} /\text{R} \text{C}})$ \Rightarrow $\frac{\text{t}}{\text{RC}}=\ln (\frac{10}{9})$ \Rightarrow $\text{C}=\frac{\text{t}}{\text{R} \ln (10 / 9)}=\frac{10^{-6}}{10 \times \ln (10 / 9)}$ $=0.95 \\mu {\text{F}}$ ∴ The capacitance of the capacitor is 0.95 μ F.

Let us consider the gravitational force acting on each mass M by adjacent particles be F. and the gravitational force acting on each mass M diagonally be $\text{F}_1$ The net force, $\text{F}_{{\text{net} }}=\frac{\text{Mv}^{2}}{\text{R}}$ Along the centre of circle, $\sqrt{2} \text{F}+\text{F}_{1}=\frac{\text{Mv}^{2}}{\text{R}}$ $\sqrt{2}(\frac{\text{GMM}}{(\sqrt{2} \text{R})^{2}})+(\frac{\text{GMM}}{(2 \text{R})^{2}})$ $=\frac{\text{Mv}^{2}}{\text{R}}$ \Rightarrow $\frac{\text{GM}}{\text{R}^{2}}(\frac{1}{\sqrt{2}}+\frac{1}{4})=\frac{v^{2}}{\text{R}}$ \Rightarrow $v=\frac{1}{2} \sqrt{\frac{\text{GM}(2 \sqrt{2}+1)}{\text{R}}}$ Hence, the speed of each particle is $\frac{1}{2} \sqrt{\frac{\text{GM}(2 \sqrt{2}+1)}{\text{R}}}$.

Given, electric field, $\text{E}=20 \cos (2 \times 10^{10} t-200 x) {\text{V}} / {\text{m}}$ Comparing with the standard equation, $\text{E}=\text{E}_{0} \cos (\omegat-k x) {\text{V}} / {\text{m}}$, we get Wave constant, k = 200 Angular frequency, $\omega=2 \times 10^{10} {\text{rad}} / {\text{s}}$ Speed of the wave, $v=\frac{\omega}{k}=\frac{2 \times 10^{10}}{200}=10^{{8}} {\text{m}} / {\text{s}}$ Refractive index, $\\mu =\frac{\text{C}}{\text{V}}=\frac{3 \times 10^{8}}{10^{8}}=3$ As we know the relation between the refractive index and dielectricconstant, $\\mu =\sqrt{\epsilon_{r} \\mu _{r}}$ Substituting the value in the above equations, we get $3=\sqrt{{\epsilon_r}(1)}$ $\epsilon_r=9$ Thus, the dielectric constant of the medium is 9.

The diagram is given below, The magnetic field due to the wire 1, $\text{B}_{{\text{wire} } 1}=\frac{\\mu _{0} \text{I}}{4 \piy}(\sin 90°+\sin \theta_{1})$ $\frac{\\mu _{0} \text{I}}{4 \piy} (1+x/\sqrt{x^{2+}y^2})$ The magnetic field due to the wire 2, $\frac{\\mu _{0} \text{I}}{4 \piy}(\sin 90°+\sin \theta_{2})$ $\text{B}_{{\text{wire} } 2}=\frac{\\mu _{0} \text{I}}{4\pi x}(1+y/\sqrt{x^{2+}y^2})$ Thus, the total magnetic field, $\text{B}=\text{B}_{1}+\text{B}_{2}$ $=\frac{\\mu _{\0} \text{I}}{4 \pi}(\frac{1}{y}+{x}/{y \sqrt{x^{2}+y^{2}}}+\frac{1}{x}+{y}/{x \sqrt{x^{2}+y^{2}}})$ $=\frac{\\mu _{\0} \text{I}}{4 \pi xy}[\text{sqrtx}^{2+}y^{2+}(x+y)]$

Given, the number of diatomic moles, n = 3 mol The increase in temperature of the diatomic mole, ΔT = 40ºC Now, the degree of freedom, f = linear + rotational + no oscillation f = 3 + 2 + 0 f = 5 Change in internal energy, $\Delta\text{U}=\text{n} \text{C}_{\text{V}} \Delta\text{T}$ where, $\text{C}_{\text{V}}=\frac{\text{f}}{2} \text{R}=\frac{5}{2} \text{R}$ Substituting the value in Eq. (i), we get $\Delta\text{U}=\frac{5 \text{R}}{2} \text{n} \Delta\text{T}$ Now, work done by the gas for isobaric process, $\text{W}=\text{p} \Delta\text{V}=\text{n} \text{R} \Delta\text{T}$ The ratio of the change in internal energy to the work done by thegas, $\frac{\Delta\text{U}}{\text{W}}=\frac{\frac{5}{2} \text{n} \text{R} \Delta\text{T}}{\text{n} \text{R} \Delta\text{T}}$ $\frac{\Delta\text{U}}{\text{W}}=\frac{5}{2}$ Multiply and divide the above equation with 5, we get $\frac{\Delta\text{U}}{\text{W}}=\frac{5 \times 5}{2 \times 5}=\frac{25}{10}$ Comparing with given equation, $\frac{\Delta\text{U}}{\text{W}}=\frac{x}{10}$ The value of the x = 25.

Given, $\text{b}_{1}=3 \text{b}_{2}$ Here, $b_1$ = width of the one of the two slit and $b_2$ = width of the other slit. As we know that, Intensity, I $\propto (\text{Amplitude}) ^{2}$ \Rightarrow $\frac{\text{I}_{1}}{\text{I}_{2}}=(\frac{b_{1}}{b_{2}})^{2}$ \Rightarrow $\frac{\text{I}_{1}}{\text{I}_{2}}=(\frac{3 b_{2}}{b_{2}})^{2}$ As we know, the ratio of the minimum intensity to the maximumintensity in the interference pattern, $\frac{\text{I}_{\min }}{\text{I}_{\max }}=({\sqrt{\text{I}_{1}}-\sqrt{\text{I}_{2}}}/{\sqrt{\text{I}_{1}}+\sqrt{\text{I}_{2}}})^{2}$ Substituting the values in the above equations, we get $\frac{\text{I}_{\min }}{\text{I}_{\max }}={(\sqrt{9 \text{I}_{2}}-\sqrt{\text{I}_{2}})^{2}}/{(\sqrt{9 \text{I}_{2}}+\sqrt{\text{I}_{2}})^{2}}=\sqrt(\frac{3-1}{3+1})^{2}$ $\frac{\text{I}_{\min }}{\text{I}_{\max }}=\frac{1}{4}$ Comparing with, $\frac{\text{I}_{\min }}{\text{I}_{\max }}=\frac{x}{4}$ The value of x = 1.

Given, the time period of the first satellite, $\text{T}_1$ = 1h The time period of the second satellite, $\text{T}_2$ = 8 h The radius of the orbit of nearer satellite, $\text{R}_1$ = 2000 kmApplying the Kepler's third law, $\text{T}^{2} \propto\text{R}^{3}$ $(\frac{\text{T}_{1}}{\text{T}_{2}})^{2}=(\frac{\text{R}_{1}}{\text{R}_{2}})^{3}$ \Rightarrow $\frac{2000}{\text{R}_{2}}=(\frac{1}{8})^{2 / 3}$ \Rightarrow $\text{R}_2$ = 8000 km As we know the relation between the angular speed and timeperiod $\omega=\frac{\theta}{\text{T}}$ So, the angular speed of the nearer satellite to the orbit, $\omega_{1}=\frac{2 \pi}{1} {\text{rad}} / {\text{h}}$ and the angular speed of the farther satellite to the orbit, $\omega_{2}=\frac{2 \pi}{8}=\frac{\pi}{4} {\text{rad}} / {\text{h}}$ The speed of the nearer satellite to orbit, $v_{1}=\omega_{1} \text{R}_{1}=(2 \pi) \times 2 \times 10^{3} {\text{km}} / {\text{h}}$ The speed of the farther satellite to the orbit $\text{V}_{2}=\omega_{2} \text{R}_{2}$ $=(\frac{\pi}{4}) \times 8000$ $=\pi\\times 2 \times 10^{3} {\text{km}} / {\text{h}}$ Thus, the relative angular speed of the nearer satellite to the farthersatellite, $\omega=\frac{\text{V}_{1}-\text{V}_{2}}{\text{R}_{1}-\text{R}_{2}}$ $=\frac{2 \pi\times 2 \times 10^{3}-\pi\times 2 \times 10^{3}}{8000-2000}$ $=\frac{2 \pi\times 10^{3}}{6000}=\frac{\pi}{3} {\text{rad}} / {\text{h}}$ Comparing with, $\omega=\frac{\pi}{x}$ The value of x = 3.

Let's draw the free body diagram when body slides down on smooth surface For smooth surface, ma = mg sin30º a = g sin30º a = g / 2 Distance covered by the block on the smooth surface in time T, $\text{s}=\text{u} \text{t}+\frac{1}{2} \text{a} \text{t}^{2}$ $\text{s}=0+\frac{1}{2}(\frac{\text{g}}{2}) \text{T}^{2}$ $\Rightarrow \text{s}=(\frac{g}{4}) \text{T}^{2}$ ...(i) Now, let's draw the free body diagram when body slides down on rough surface For rough surface, ma = mg sin30º - μ mg cos30º a = g sin30º - μ g Distance covered by the block on the rough surface in time alpha T, $\text{s}=\text{u} \text{t}+\frac{1}{2} \text{a} \text{t}^{2}$ $\text{s}=0+\frac{1}{2}(\text{g} \sin 30\º - \\mu \text{g} \sin 30\\mu ) \text{t}^{2}$ $\text{s}=\frac{\text{g}}{4}(1-\sqrt{3} \\mu )(\alpha\text{T})^{2}$ ...(ii) Distance covered by the block is same for both the case, $\Rightarrow \frac{\text{g}}{4}(1-\sqrt{3} \\mu )(\alpha\text{T})^{2}=\frac{\text{g}}{4} \text{T}^{2}$[from Eq. (i) & Eq. (ii)] $\Rightarrow 1-\sqrt{3} \\mu =\frac{1}{\alpha^{2}}$ $\Rightarrow \\mu =(\frac{\alpha^{2}-1}{\alpha^{2}}) \frac{1}{\sqrt{3}}$ Comparing with $(\frac{\alpha^{2}-1}{\alpha^{2}}) {1}/{\sqrt{\text{x}}}$ The value of the x = 3.

Given, the average translational kinetic energy of dinitrogen(\text{N}_2$) = Kinetic energy of an electron ...(i) Translational kinetic energy of dinitrogen (\text{N}_2$ ) ${\text{KE}}=\frac{3}{2} \text{K}_{\text{B}} \text{T}$ Here, T = temperature of the gas, and $\text{K}_\text{B}$ = Boltzmann constant. Kinetic energy of an electron = eV Given, the potential difference of an electron, V = 0.1 V Substituting the values in the Eq. (i), we get $\frac{3}{2} \text{K}_{\text{B}}\ \text{T}={\text{eV}}$ \Rightarrow $\frac{3}{2} \times 1.38 \times 10^{-23} \times \text{T}=1.6 \times 10^{-19} \times (0.1)$ $\text{T}=773 {\text{K}}=773-273° {\text{C}}=500° {\text{C}}$

For case I, The potential difference of the uniform wire, V =240 V The resistance of the uniform wire, $\text{R}_1$ = 36 Ω The power dissipation in the first case, $\text{P}_{1}=\frac{\text{V}^{2}}{\text{R}_{1}}=\frac{(240)^{2}}{36}$ For case II, The resistance of each half, $\text{R}_{2}=\frac{\text{R}_{1}}{2}=\frac{36}{2}=18 \Omega$ $\text{P}_{2}=\frac{\text{V}^{2}}{\text{R}_{2}}+\frac{\text{V}^{2}}{\text{R}_{2}}$ $=\frac{(240)^{2}}{18}+\frac{(240)^{2}}{18}=\frac{(240)^{2}}{9}$ Thus, the ratio of the total power dissipation in the first case to thesecond case $\frac{\text{P}_{1}}{\text{P}_{2}}=\frac{(240)^{2} / 36}{(240)^{2} / 9}$ \Rightarrow $\frac{\text{P}_{1}}{\text{P}_{2}}=\frac{1}{4}$ Comparing with, $\frac{\text{P}_{1}}{\text{P}_{2}}=\frac{1}{x}$ The value of the x = 4.

Given, the Young's modulus of the steel rod, Y = 2 × $10^11$ Pa Thermal coefficient of the steel rod, alpha = $10^{-5}$ºC The length of the steel rod, I = 4 m The area of the cross-section, A = 10 $\text{cm}^2$ The temperature difference, ΔT = 400°C As we know that, Thermal strain = alpha ΔT Using the Hooke's law Young's modulus $(\text{Y})={{ \text{Thermal} \text{stress} }}/{{ \text{Thermal} \text{strain} }}=\frac{\text{F} / \text{A}}{\alpha \Delta\text{T}}$ Thermal stress, F = YA alpha Δ T Substitute the values in the above equation, we get $\text{F}=2 \times 10^{11} \times 10 \times 10^{-4} \times 10^{-5} \times (400)$ $=8 \times 10^{5} {\text{N}}$ Comparing with, $\text{F}=x \times 10^{5} {\text{N}}$ The value of the x = 8.

Given, the mass of the steel rod, m = 2 kg The length of the steel rod, I = 0.6 m According to question, the diagram can be as shown below By using the energy conservation, $m g l=\frac{1}{2} \text{l} \omega^{2}$ $m g l=\frac{1}{2} \times \frac{m l^{2}}{3} \times \omega^{2}$ \Rightarrow $\omega = \frac{\sqrt6g}{\text{l}}$ As we know the relation between the linear speed and angularspeed, $v=\omega r=\omegal=\sqrt{6 g l}$ Substituting the values in the above equation, we get $\text{v}=\sqrt{6 \times 10 \times 0.6}=6$ m / s Hence, the speed of the free end of the rod when it passes throughits lowest position is 6 m/s.

Given, the amplitude of the carrier wave, $\text{A}_\text{c}$ = 250 V The amplitude of the message wave, $\text{A}_\text{m}$ = 150 V We know that, The maximum amplitude, \text{A}_\max = $\text{A}_\text{c} + \text{A}_\text{m}$ Substituting the values in the above equation, we get \text{A}_\max = 250 + 150 = 400 V We know that, The minimum amplitude, \text{A}_\min = \text{A}_\text{c} - \text{A}_\text{m} Substituting the values in the above equation, we get \text{A}_\min = 250 - 150 = 100 V Thus, the ratio of the minimum amplitude to the maximumamplitude of the modulated wave is $\frac{\text{A}_{\min }}{\text{A}_{\max }}=\frac{100}{400}=\frac{1}{4}$ Comparing with, 1: x The value of the x = 4.

Given, the length of the shock absorber, I = 1.5 m The total mass of the system, M = 40000 kg The speed of the wagon, v = 72 km/h When brakes are applied, the final velocity, $\text{v}_\text{f}$ = 0 The compressed spring of the shock absorber, x = 1mApplying the work-energy theorem, Work done by the system = Change in kinetic energy W = Δ KE $\text{W}_{{\text{friction} }}+\text{W}_{{\text{spring} }}=\frac{1}{2} \text{m} \text{v}_{\text{f}}^{2}+\frac{1}{2} \text{m} \text{v}_{\text{i}}^{2}$ $-\frac{90}{100}(\frac{1}{2} \text{m} \text{v}^{2})+\text{W}_{{\text{spring} }}=0-\frac{1}{2} \text{m} \text{v}_{\text{i}}^{2}$ (\because 90% energy lost due to friction) $\text{W}_{{\text{spring} }}=-\frac{10}{100} \times \frac{1}{2} \text{m} \text{v}^{2}$ $-\frac{1}{2} \text{k} \text{x}^{2}=\frac{1}{20} \text{m} \text{v}^{2}$ $\text{k}=\frac{\text{m} \text{v}^{2}}{10 \times x^{2}}$ Substituting the values in the above equation, we get $\text{k}=\frac{40000 \times (72 \times \frac{5}{18})^{2}}{10(1)^{2}}$ = 16 × $10^5$ N / m Comparing the spring constant, k = x × $10^5$ The value of the x = 16.

BOD is the biological oxygen demand. Cleanest water sample will have BOD value equal to (\text{SO}_2$ ) 3 ppm as clean water could have BOD value of less than 5 ppm.

Calamine is the ore of Zn and is found as zinc carbonate (\text{Zn}\text{CO}3$), where malachite is the ore of copper and is found as copper hydroxide [Cu$(\text{OH}) 2$] and copper carbonate [CuCO3].

A functional group cannot be reduced by $\text{Na}/ \text{H}_2$ as it is not a reducing agent.