31 Jan 2024 Shift 2

We need to find $|\vec{A} + \vec{B}|$ for two vectors of equal magnitude $R$ inclined at angle $\theta$. Since the magnitude of the \sum of two vectors is given by the parallelogram law, $|\vec{A} + \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta$. Substituting $|\vec{A}| = |\vec{B}| = R$ into this expression gives $|\vec{A} + \vec{B}|^2 = R^2 + R^2 + 2R^2\cos\theta = 2R^2(1 + \cos\theta)$. Using the double angle identity $1 + \cos\theta = 2\cos^2\left(\frac{\theta}{2}\right)$, it follows that $|\vec{A} + \vec{B}|^2 = 2R^2 \times 2\cos^2\left(\frac{\theta}{2}\right) = 4R^2\cos^2\left(\frac{\theta}{2}\right)$. Now, taking the positive square root yields $|\vec{A} + \vec{B}| = 2R\cos\left(\frac{\theta}{2}\right)$. For verification, one can similarly compute $|\vec{A} - \vec{B}|^2 = 2R^2(1-\cos\theta) = 4R^2\sin^2\left(\frac{\theta}{2}\right)$, giving $|\vec{A}-\vec{B}| = 2R\sin\left(\frac{\theta}{2}\right)$, which matches Option 1's formula but with coefficient 2 (not $\sqrt{2}$) and therefore Option 1 is incorrect. The correct answer is Option 3 : $|\vec{A} + \vec{B}| = 2R\cos\frac{\theta}{2}$.

$E=\frac{B-x^2}{At}$. $[B]=\$[x^2]\$=L^2$. $[A]=[B]/([E][t])=L^2/(ML^2T^{-2}\cdot T)=L^2/(ML^2T^{-1})=M^{-1}T$. $[AB]=M^{-1}T\cdot L^2=L^2M^{-1}T^1$. The answer is Option (2): $L^2M^{-1}T^1$.

Lets start writing equations for both the blocks For $m_1$ : - $m_1g-T=m_1a$ For $m_2$: - $T-m_2g=m_2a$ Given that $a=\frac{g}{8}$ On substituting a \in both equations and adding them we get , $m_1g-\ m_2g\ =\ m_1\left(\frac{g}{8}\right)+m_2\left(\frac{g}{8}\right)$ $\left(m_1-\ m_2\right)g\ =\ \left(m_1+m_2\right)\left(\frac{g}{8}\right)$ $\frac{\left(m_1-\ m_2\right)}{m_1+m_2}=\frac{1}{8}$ On dividing Numerator and denominator with $m_2$ $\frac{\left(\frac{m_1}{m_2}-1\right)}{\left(\frac{m_1}{m_2}+1\right)}=\frac{1}{8}$ consider $\frac{m_1}{m_2}$ as x and solve for x $\frac{\left(x-1\right)}{\left(x+1\right)}=\frac{1}{8}$ $\left(x-1\right)=\frac{1}{8}\left(x+1\right)$ $x-\frac{1}{8}x=\frac{1}{8}+1$ $\frac{7}{8}x=\frac{9}{8}$ $x=\frac{9}{7}$

$m=5kg,\theta =30°,\mu =0.1=1/10,g=10$ Normal reaction: $N=mg\cos\theta =5\times10\times \cos30°=50\times (\sqrt{3}/2)=25\sqrt{3}$ Friction: $f=\mu N=(1/10)\times25\sqrt{3}=(5\sqrt{3})/2$ Now case 1: just moving up the plane Friction acts down the plane $F₁=mg\sin\theta +f$ $=5\times10\times (1/2)+(5\sqrt{3})/2$ $=25+(5\sqrt{3})/2$ Now case 2: just preventing sliding down Friction acts up the plane $F₂=mg\sin\theta −f$ $=25−(5\sqrt{3})/2$ Now difference: $|F₁|−|F₂|$ $=[25+(5\sqrt{3})/2]−[25−(5\sqrt{3})/2]$ $\frac{5\sqrt{3}}{2}-\left(-\frac{5\sqrt{3}}{2}\right)=\frac{5\sqrt{3}}{2}+\frac{5\sqrt{3}}{2}$ $=5\sqrt{3}$ Final answer: $=5\sqrt{3}$

$\vec{F}=(6t)\hat{i}+(6t^2)\hat{j}$. $m=2$ kg. $\vec{a}=\vec{F}/m=(3t)\hat{i}+(3t^2)\hat{j}$. $\vec{v}=\int\vec{a}dt=\frac{3t^2}{2}\hat{i}+t^3\hat{j}$. Power: $P=\vec{F}\cdot\vec{v}=(6t)(3t^2/2)+(6t^2)(t^3)=9t^{3+}6t^5$. The answer is Option (4): $(9t^{3+}6t^5)$ W.

We need to find the escape velocity on the moon relative to the planet. Since $M_m = \frac{M_p}{144}$, $d_m = \frac{d_p}{16}$, we have $R_m = \frac{R_p}{16}$ and the escape velocity on the planet is $v$. We start with the escape velocity formula: $v_e = \sqrt{\frac{2GM}{R}}$ Next, the ratio of the escape velocity on the moon to that on the planet is given by $\frac{v_m}{v_p} = \sqrt{\frac{M_m/R_m}{M_p/R_p}} = \sqrt{\frac{M_m \cdot R_p}{M_p \cdot R_m}}$ Substituting the expressions for mass and radius yields $= \sqrt{\frac{1}{144} \times 16} = \sqrt{\frac{16}{144}} = \frac{4}{12} = \frac{1}{3}$ This gives $v_m = \frac{v}{3}$. The correct answer is Option 1 : $\frac{v}{3}$.

We need to find the terminal velocity of a ball of radius $2r$ with the same mass as a ball of radius $r$. According to Stokes' law, the terminal velocity is given by $v_t = \frac{2r^2(\rho_s - \rho_l)g}{9\eta}$. Since the medium has negligible density ($\rho_l \approx 0$), this reduces to $v_t = \frac{2r^2\rho_s g}{9\eta}$. Because the spheres have the same mass, we have $\frac{4}{3}\pi r_1^3 \rho_1 = \frac{4}{3}\pi r_2^3 \rho_2$ with $r_1 = r$ and $r_2 = 2r$. Substituting these radii gives $\rho_2 = \rho_1 \left(\frac{r_1}{r_2}\right)^3 = \rho_1 \left(\frac{r}{2r}\right)^3 = \frac{\rho_1}{8}$. Then the ratio of their terminal velocities is $\frac{v_2}{v_1} = \frac{r_2^2 \rho_2}{r_1^2 \rho_1} = \frac{(2r)^2 \times \rho_1/8}{r^2 \times \rho_1} = \frac{4r^2 \times \rho_1/8}{r^2 \times \rho_1} = \frac{4}{8} = \frac{1}{2}$. Hence $v_2 = \frac{v}{2}$. The correct answer is Option 1 : $\frac{v}{2}$.

8 mol Ar (monoatomic): $U_1=8\cdot\frac{3}{2}RT=12RT$. 6 mol O₂ (diatomic, no vibration): $U_2=6\cdot\frac{5}{2}RT=15RT$. Total: $U=12RT+15RT=27RT$. The answer is Option (3): $27RT$.

We need to find the speed of sound in oxygen at STP. The values of the gas constant and the adiabatic index are given as $R = 8.3$ J/(mol K) and $\gamma = 1.4$, respectively, while at STP the temperature is $T = 273$ K and the molar mass of $O_2$ is 32 g/mol = 0.032 kg/mol. The speed of sound \in a gas is given by the formula $v = \sqrt{\frac{\gamma RT}{M}}$. Substituting \in the numbers leads to $v = \sqrt{\frac{1.4 \times 8.3 \times 273}{0.032}}$. Since the numerator is $1.4 \times 8.3 = 11.62$, multiplying this by 273 gives $11.62 \times 273 = 3172.26$. This gives $v = \sqrt{\frac{3172.26}{0.032}} = \sqrt{99133} \approx 315 \text{ m/s} \approx 310 \text{ m/s}$. Therefore the correct answer is Option 1 : 310 m/s.

We need to find the force between charges in a dielectric medium at a reduced distance. In vacuum at distance $r$ the force is given by $F = \frac{kq_1q_2}{r^2}$. Since the dielectric constant is $K = 5$ and the separation is reduced to $r/5$, Coulomb's law \in the medium yields: $F' = \frac{kq_1q_2}{K \cdot (r/5)^2} = \frac{kq_1q_2}{5 \cdot r^2/25} = \frac{25kq_1q_2}{5r^2} = \frac{5kq_1q_2}{r^2} = 5F$ This shows that although the dielectric reduces the force by a factor of $K = 5$, the shorter separation increases it by a factor of 25, resulting \in a net factor of $\frac{25}{5} = 5$ \times the original force. Therefore, the correct answer is Option 2 : $5F$.

We need to find the percentage decrease in illumination when current drops by 20%. Since for a resistive lamp the power (and hence illumination) is proportional to $I^2$, we have $P = I^2 R$. If the current drops by 20%, the new current becomes $I' = 0.8I$ and substituting into the expression for power gives $P' = (0.8I)^2 R = 0.64 I^2 R = 0.64P$. Therefore, the percentage decrease is given by $\frac{P - P'}{P} \times 100 = (1 - 0.64) \times 100 = 36\%$. The correct answer is Option 3 : 36%.

We need to determine the new balancing length in a meter bridge when the wire is replaced. The balance condition in a meter bridge is $\frac{X}{R} = \frac{R_l}{R_{100-l}}$ where $R_l$ and $R_{100-l}$ are the resistances of the wire on the two sides. Since the resistance of a uniform wire is proportional to its length, this becomes $\frac{X}{R} = \frac{r \cdot l}{r \cdot (100-l)} = \frac{l}{100-l}$ and hence the resistance per cm ($r$) cancels out. Substituting the original values gives $\frac{X}{25} = \frac{40}{60} = \frac{2}{3}$, so $X = \frac{50}{3}$ ohm. When the wire is replaced by one with resistance $2r$ per cm, we have $\frac{X}{25} = \frac{2r \cdot l'}{2r \cdot (100-l')} = \frac{l'}{100-l'}$. The $2r$ cancels from both sides, giving the same equation as before. Therefore, the balancing length remains $l' = 40$ cm. The correct answer is Option 4 : 40 cm.

Magnetic field $B = 2 \times 10^{-3}$ T along +Y. Rectangular loop 20 cm \times 10 cm \in Y-Z plane, current 5 A anticlockwise w.r.t. negative X-axis. The area vector of the loop \in Y-Z plane points along $\pm\hat{x}$. Since current is anticlockwise when viewed from negative X-axis, by the right-hand rule, the area vector points along $-\hat{x}$. $\vec{m} = I\vec{A} = 5 \times (20 \times 10 \times 10^{-4})(-\hat{x}) = 5 \times 0.02(-\hat{x}) = -0.1\hat{x}$ A m$^2$. Torque: $\vec{\tau} = \vec{m} \times \vec{B} = (-0.1\hat{x}) \times (2 \times 10^{-3}\hat{y})$ $= -0.1 \times 2 \times 10^{-3}(\hat{x} \times \hat{y}) = -2 \times 10^{-4}\hat{z}$ Magnitude: $2 \times 10^{-4}$ N m, direction: along negative Z-direction. The correct answer is Option B: $2 \times 10^{-4}$ N m along negative Z-direction.

$V_0=20$V, $I_0=10$A. Phase difference $\phi=\pi/3$. $P_{avg}=\frac{V_0I_0}{2}\cos\phi=\frac{20\times10}{2}\cos60°=100\times0.5=50$ W. The answer is Option (4): 50 W.

We need to evaluate two statements about electromagnetic waves. Analysis of Statement I: "Electromagnetic waves carry energy as they travel through space and this energy is equally shared by the electric and magnetic fields." An electromagnetic wave has energy density given by: $u = \frac{1}{2}\epsilon_0 E^2 + \frac{B^2}{2\mu_0}$ For an EM wave, the relationship $E = cB$ holds, where $c = 1/\sqrt{\mu_0 \epsilon_0}$. Substituting: $\frac{1}{2}\epsilon_0 E^2 = \frac{1}{2}\epsilon_0 c^2 B^2 = \frac{B^2}{2\mu_0}$ This shows that the electric field energy density equals the magnetic field energy density. The energy is indeed equally shared between the two fields. Statement I is CORRECT. Analysis of Statement II: "When electromagnetic waves strike a surface, a pressure is exerted on the surface." Electromagnetic waves carry momentum $p = E/c$ (where $E$ is energy). When they strike a surface, they transfer momentum, exerting radiation pressure. For a perfectly absorbing surface, the radiation pressure is $P = I/c$, where $I$ is the intensity. For a perfectly reflecting surface, $P = 2I/c$. Statement II is CORRECT. The correct answer is Option 2 : Both Statement I and Statement II are correct.

We need to find the angle of refraction when unpolarised light is reflected with complete polarisation. Key Concept: Brewster's Law When unpolarised light strikes a surface at the Brewster angle ($i_B$), the reflected light is completely polarised. At this angle: $\tan i_B = \mu$ Also, at Brewster's angle, the reflected and refracted rays are perpendicular: $i_B + r = 90°$ Angle of incidence = 60° (this is the Brewster angle). Finding the angle of refraction: $r = 90° - i_B = 90° - 60° = 30°$ Verification: $\tan 60° = \sqrt{3} = \mu$. By Snell's law: $\sin 60° = \sqrt{3} \sin r$, so $\sin r = \frac{\sqrt{3}/2}{\sqrt{3}} = \frac{1}{2}$, giving $r = 30°$. ✓ The correct answer is Option 1 : 30°.

We need to determine what happens to photoelectron emission when frequency is halved and intensity is doubled. Key Concept: Photoelectric Effect Photoelectrons are emitted only when the frequency of incident light is greater than or equal to the threshold frequency ($\nu_0$). This is Einstein's photoelectric equation: $KE_{max} = h\nu - h\nu_0$ Initial condition: Frequency = $1.5\nu_0 > \nu_0$ (above threshold, photoelectrons are emitted). New condition: Frequency halved = $\frac{1.5\nu_0}{2} = 0.75\nu_0$. Since $0.75\nu_0 < \nu_0$ (below threshold frequency), no photoelectrons will be emitted , regardless of the intensity. Increasing intensity only increases the number of photons, but each photon still has energy $h \times 0.75\nu_0 < h\nu_0$, which is insufficient to eject an electron. The correct answer is Option 3 : Zero.

We need to find the mass number of a nucleus whose radius is half that of a nucleus with mass number 192. Nuclear Radius Formula: $R = R_0 A^{1/3}$ where $R_0 \approx 1.2$ fm is a constant and $A$ is the mass number. Setting up the equation: $R_2 = \frac{R_1}{2}$, so $R_0 A_2^{1/3} = \frac{R_0 A_1^{1/3}}{2}$. $A_2^{1/3} = \frac{A_1^{1/3}}{2}$ Cubing both sides: $A_2 = \frac{A_1}{8} = \frac{192}{8} = 24$ The correct answer is Option 1 : 24.

The output of the top OR gate with inputs A and $\bar{B}$ is $A + \bar{B}$. The output of the bottom OR gate with inputs $\bar{A}$ and B is $\bar{A} + B$. These two signals are inputs to the final NOR gate. The expression for the final output Y is $Y = \overline{(A + \bar{B}) + (\bar{A} + B)}$. Let's simplify the expression inside the overbar: $(A + \bar{B}) + (\bar{A} + B) = (A + \bar{A}) + (B + \bar{B})$. According to the laws of Boolean algebra, $X + \bar{X} = 1$. Therefore, the expression simplifies to $1 + 1 = 1$. The final output is then $Y = \bar{1} = 0$. This indicates that the output Y is always 0 for all possible input combinations of A and B, which corresponds to the truth table in option C.

We need to find the accuracy (percentage error) in the measurement of $g$ using a simple pendulum. Length: $l = 20$ cm, $\Delta l = 2$ mm = 0.2 cm Time for 50 oscillations: $t = 40$ s, $\Delta t = 1$ s From $T = 2\pi\sqrt{l/g}$, we get: $g = \frac{4\pi^2 l}{T^2}$ Since $T = t/50$: $g = \frac{4\pi^2 l \times 2500}{t^2} = \frac{10000\pi^2 l}{t^2}$ Taking logarithmic differentiation: $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T} = \frac{\Delta l}{l} + 2\frac{\Delta t}{t}$ (Since $T = t/50$, $\Delta T/T = \Delta t/t$.) $\frac{\Delta g}{g} = \frac{0.2}{20} + 2 \times \frac{1}{40} = 0.01 + 0.05 = 0.06 = 6\%$ Therefore $N = 6$. The correct answer is Option 3 : 6.

Two spheres: each mass 2kg, radius 0.5m, separation 1.5m (center to center), so each is 0.75m from middle. $I=2\left[\frac{2}{5}mr^{2+}md^2\right]\$=2\left[\frac{2}{5}(2)(0.25)+(2)(0.5625)\right]=2[0.2+1.125]=2(1.325)=2.65=\frac{53}{20}$ kg m². So $x=53$. The answer is $\boxed{53}$.

given mass = m projection speed = u angle = 45° we need angular momentum about point of projection at highest point formula: L = m | r × v | for 2D motion: $L=m(xv_y−yv_x)$ step 1: velocity at highest point at highest point vertical velocity = 0 horizontal velocity = $u\cos45°=u/\sqrt{2}$ so: v_x = u/√2 , v_y = 0 step 2: time to reach highest point $t=(u\sin45°)/g$ = (u/√2)/g = u/(√2 g) step 3: coordinates of highest point horizontal distance: $x=(u\cos45°)\times t$ = (u/√2) \times (u/(√2 g)) = u²/(2g) vertical height: $y=(u\sin45°)^2/(2g)$ = (u²/2)/(2g) = u²/(4g) step 4: angular momentum $L=m(xv_y−yv_x)$ $\sin ce\ v_y=0:$ $L=−myv_x$ magnitude: $L=myv_x$ $=m\times (u^2/(4g))\times (u/\sqrt{2})$ $=\mu^3/(4\sqrt{2}g)$ Now comparing with the given expression :- $\frac{\sqrt{2}\mu^3}{Xg}$ we get it by multiplying and dividing our answer by $\sqrt{\ 2}$ so we get $=\sqrt{\ 2}\mu^3/8g$ so Value of X is 8

Masses$:m₁=2kg,m₂=4kg$ For Atwood machine: Acceleration: $a=\frac{(m₂−m₁)g}{(m₁+m₂)}$ = $\frac{\left((4−2)\times10\right)}{(2+4)}$ $=\frac{20}{6}=\frac{10}{3}$ Tension: $T=m₁(g+a)$ $=2(10+10/3)$ =$2\times (40/3)$ $=\frac{80}{3}N$ Now strain: $strain=stress/Y$ $Stress=T/A$ Area of wire: $A=\pi r^2=\pi(4\times10⁻⁵)^2$ =\pi\times16\times10⁻^1⁰ =16\pi\times10⁻^1⁰ So: $strain=T/(AY)$ =(80/3)/[(16\pi\times10⁻^1⁰)(2\times10^{11})] Simplify denominator: $16\times2=32$ 10⁻^1⁰\times10^{11}=10 $Sodeno\min ator=32\pi\times10=320\pi$ Now: $strain=(80/3)/(320\pi)$ $=80/(960\pi)$ $=1/(12\pi)$ So given: $strain=1/(\alpha \pi)$ $\Rightarrow \alpha =12$ Final answer: $12$

We need equivalent spring constant first. Left spring has spring constant k On right side, the two upper springs (k and k) are in parallel. So their equivalent is $k_p=k+k=2k$ This combination is \in series with lower spring k. So right-side equivalent: $\frac{1}{k_r}=\frac{1}{2k}+\frac{1}{k}$ =\frac{1+2}{2k}$$$$=\frac{3}{2k} Thus $k_r=\frac{2k}{3}$ Now this right equivalent acts \in parallel with left spring k. Total effective spring constant: $k_{\text{eff}}=k+\frac{2k}{3}$ $=\frac{5k}{3}$ Time period: $T=2\pi\sqrt{\frac{M}{k_{\text{eff}}}}$ $2\pi\sqrt{\frac{3M}{5k}}$ $\pi\sqrt{\frac{12M}{5k}}$ Given $T=\pi\sqrt{\frac{\alpha M}{5k}}$ Comparing, $\alpha =12$

step 1: positions of charges On the x-axis: • +2q at −2l • −q at −l • +q at +l • −2q at +2l step 2: net charge $2q-q+q-2q=0$ So monopole term = 0 step 3: dipole moment Take dipole moment about O: $p=\sum_{ }^{ }q_ix_i$ $=(2q)(-2l)+(-q)(-l)+(q)(l)+(-2q)(2l)$ $=-4ql+ql+ql-4ql=-6ql$ So, $p=-6ql$ (direction along negative x-axis) step 4: potential at point P For dipole: $V=\frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^2}$ Here $\theta=60^{\circ}$ $\cos60^{\circ}=\frac{1}{2}$ $V=9\times10^9\cdot\frac{-6ql\cdot(1/2)}{r^2}$ $=9\times10^9\cdot\frac{-3ql}{r^2}$

The resistor network is symmetrical and can be simplified into three parallel branches. The top branch is a $2\ \Omega$ resistor, and the bottom branch is a $2\ \Omega$ resistor. The central 'X' structure is a balanced Wheatstone bridge (formed by four $2\ \Omega$ resistors), which has an upper arm of $2+2=4\ \Omega$ and a lower arm of $2+2=4\ \Omega$. Its equivalent resistance is $\frac{4 \times 4}{4+4} = 2\ \Omega$.

The total external resistance $R_{ext}$ is the parallel combination of these three branches:
$R_{ext} = \left(\frac{1}{2} + \frac{1}{2} + \frac{1}{2}\right)^{-1} = \left(\frac{3}{2}\right)^{-1} = \frac{2}{3}\ \Omega$

The total resistance of the entire circuit is the sum of the external resistance and the internal resistance $r$:
$R_{total} = R_{ext} + r = \frac{2}{3}\ \Omega + \frac{2}{3}\ \Omega = \frac{4}{3}\ \Omega$

The power consumption in the entire circuit is given by $P = \frac{E^2}{R_{total}}$:
$P = \frac{(2 \text{ V})^2}{\frac{4}{3}\ \Omega} = \frac{4}{\frac{4}{3}} = 3 \text{ W}$

A circular coil of $N$ turns and radius $R$ produces, at its centre, a magnetic field along the axis of the coil given by $B = \frac{\mu_0 N I}{2R} \quad -(1)$ where $I$ is the current through the coil. For coil $P$: Number of turns, $N_P = 100$ Current, $I_P = 1 \, \text{A}$ Radius, $R = \pi \, \text{cm} = \pi \times 10^{-2} \, \text{m}$ Substituting \in $(1)$, $B_P = \frac{\mu_0 N_P I_P}{2R}$ $= \frac{(4\pi \times 10^{-7}) (100)(1)}{2(\pi \times 10^{-2})}$ $= \frac{4\pi \times 10^{-5}}{2\pi \times 10^{-2}}$ $= 2 \times 10^{-3} \, \text{T}$ $= 2 \, \text{mT} \quad -(2)$ For coil $Q$: Number of turns, $N_Q = 100$ (same as $P$) Current, $I_Q = 2 \, \text{A}$ Again using $(1)$, $B_Q = \frac{\mu_0 N_Q I_Q}{2R}$ $= \frac{(4\pi \times 10^{-7}) (100)(2)}{2(\pi \times 10^{-2})}$ $= \frac{8\pi \times 10^{-5}}{2\pi \times 10^{-2}}$ $= 4 \times 10^{-3} \, \text{T}$ $= 4 \, \text{mT} \quad -(3)$ The planes of the two coils are mutually perpendicular, so the fields $B_P$ and $B_Q$ are perpendicular to each other. Hence, the resultant field at the common centre is obtained by vector addition using the Pythagoras theorem: $B_{\text{res}} = \sqrt{B_P^{\,2} + B_Q^{\,2}}$ $= \sqrt{(2 \, \text{mT})^{2} + (4 \, \text{mT})^{2}}$ $= \sqrt{4 + 16} \, \text{mT}$ $= \sqrt{20} \, \text{mT}$ The question states $B_{\text{res}} = \sqrt{x} \, \text{mT}$. Comparing, $x = 20$. Hence, the required value is 20 .

$\phi=5t^{2-}36t+1$. $emf=-d\phi/dt=-(10t-36)$. At $t=2$: $emf=-(20-36)=16$ V. $I=emf/R=16/8=2$ A. The answer is $\boxed{2}$.

We need to find the image distance from the object when light from a point source in air falls on a convex curved surface of glass. Refraction at a single spherical surface is governed by the formula $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$. Here $\mu_1$ is the refractive index of the medium containing the object, $\mu_2$ is the refractive index of the medium containing the image, $u$ is the object distance, $v$ is the image distance, and $R$ is the radius of curvature. We know that the refractive index of air is $\mu_1 = 1$ and that of glass is $\mu_2 = 1.5$. Since the object lies on the same side as the incident light, the object distance is $u = -100\ \text{cm}$. For a convex surface the center of curvature lies on the transmission side, so $R = +20\ \text{cm}$. Substituting these values into the formula gives $\frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20}$ which simplifies to $\frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20}\,.$ On the right-hand side we have $\frac{0.5}{20} = 0.025$ so that $\frac{1.5}{v} = 0.025 - \frac{1}{100} = 0.025 - 0.01 = 0.015$ and therefore $v = \frac{1.5}{0.015} = 100\ \text{cm}\,.$ The positive value of $v$ indicates the image is formed on the transmission side (inside the glass), 100 cm from the surface. Hence, the distance from the object to the image is $|u| + v = 100 + 100 = 200\ \text{cm}$ since they lie on opposite sides of the refracting surface. The answer is 200 cm.