Question 1:
easyThe linear momentum of a particle is given by P = a + bt², where t is time and a & b are constants. The force acting on the body is directly proportional to :
Given momentum of the object is P = a + bt²;
Force is defined as rate of change of momentum F= dP/dt= 2bt. so Force is proportional to t.
Question 2:
easyWhen an oscillator completes 100 oscillation its amplitude reduced to $\frac{1}{3}$ of initial value. What will be its amplitude, oscillation when it completes 200 oscillation? (2002)
Amplitude after $n$ oscillations is given by $A = A_0 e^{-k n}$. For $n = 100$, $A_{100} = A_0 e^{-100k} = A_0 \frac{1}{3}$. For $n = 200$, $A_{200} = A_0 e^{-200k} = A_0 (e^{-100k})^2 = A_0 (\frac{1}{3})^2 = A_0 \frac{1}{9}$.
Question 3:
easyTwo hollow conducting sphere of radii $R_1$ and $R_2$ ($R_1 \gg R_2$) have equal charges. The potential would be: (2022)
Potential on the surface of a charged conducting sphere is $V = \frac{kQ}{R}$. For equal charges, $V \propto \frac{1}{R}$. Since $R_2 < R_1$, the potential is more on the smaller sphere.
Question 4:
easyTwenty seven drops of same size are charged at $220\text{ V}$ each. They combine to form a bigger drop. Calculate the potential of the bigger drop. (2021)
By conservation of volume, $27 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \implies R = 3r$. Total charge $Q = 27q$. The new potential is $V_{new} = \frac{kQ}{R} = \frac{k(27q)}{3r} = 9V_{old}$. Thus, $V_{new} = 9 \times 220\text{ V} = 1980\text{ V}$.
Question 5:
easyA short electric dipole has a dipole moment of $16 \times 10^{-9}\text{ C m}$. The electric potential due to the dipole at a point at a distance of $0.6\text{ m}$ from the centre of the dipole, situated on a line making an angle of $60^{\circ}$ with the dipole axis is : $\left(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\text{ N m}^2/\text{C}^2\right)$ (2020)
Electric potential due to a short dipole is $V = \frac{kp\cos\theta}{r^2}$. Substituting the values: $V = \frac{9 \times 10^9 \times 16 \times 10^{-9} \times \cos(60^{\circ})}{(0.6)^2} = \frac{144 \times 0.5}{0.36} = 200\text{ V}$.
Question 6:
easyFour point charges $-Q$, $-q$, $2q$ and $2Q$ are placed, one at each corner of the square. The relation between $Q$ and $q$ for which the potential at the center of the square is zero is: (2012 Pre)
The potential at the center is the scalar sum of potentials due to all four charges. Let the distance to the center be $r$. $V = \frac{k}{r}(-Q - q + 2q + 2Q) = \frac{k}{r}(Q + q)$. For $V = 0$, $Q + q = 0 \implies Q = -q$.
Question 7:
easyThree concentric spherical shells have radii $a$, $b$ and $c$ ($a < b < c$) and have surface charge densities $\sigma$, $-\sigma$ and $\sigma$ respectively. If $V_A$, $V_B$ and $V_C$ denote the potentials of the three shells, then for $c = a + b$, we have: (2009)
Potential of shell A is $V_A = \frac{\sigma}{\epsilon_0}(a - b + c)$. Potential of shell C is $V_C = \frac{\sigma}{\epsilon_0}\left(\frac{a^2 - b^2}{c} + c\right)$. Given $c = a + b$, $V_C = \frac{\sigma}{\epsilon_0}\left(\frac{(a-b)(a+b)}{a+b} + c\right) = \frac{\sigma}{\epsilon_0}(a - b + c) = V_A$. Thus $V_A = V_C \neq V_B$.
Question 8:
easySome charge is being given to a conductor. Then its potential: (2002)
In electrostatic equilibrium, the electric field inside a conductor is zero ($E = 0$). Since $E = -\frac{dV}{dr}$, the potential $V$ must be constant everywhere inside and on the surface of the conductor.
Question 9:
easyEight equally charged tiny drops are combined to form a big drop. If the potential on each drop is $10 \text{ V}$ then potential of big drop will be: (1999)
By conservation of volume, $8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \implies R = 2r$. Charge is conserved, so $Q = 8q$. The new potential $V' = \frac{kQ}{R} = \frac{k(8q)}{2r} = 4 \left(\frac{kq}{r}\right) = 4 \times 10 \text{ V} = 40 \text{ V}$.
Question 10:
easyThere is an electric field $E$ in x-direction. If the work done in moving a charge of $0.2 \text{ C}$ through a distance of $2 \text{ m}$ along a line making an angle $60^{\circ}$ with x-axis is $4 \text{ J}$, then what is the value of $E$? (1995)
Work done is given by $W = qEd \cos\theta$. Substituting the values: $4 = 0.2 \times E \times 2 \times \cos(60^{\circ})$. This simplifies to $4 = 0.4 \times E \times 0.5$, which gives $E = \frac{4}{0.2} = 20 \text{ N/C}$.