Given,  W_AB = 100 J,  q = 4C,  V_A =- 10 V,  VB = V = ?
 Since,  W_AB  = q(V_B - V_A )
\(\Rightarrow \) 100 = 4(V +10)   \(\Rightarrow \) V = 15 V

Only (c) is right; the rest cannot represent electrostatic field lines.
(a) is wrong because field lines must be normal to a conductor.
(b) is wrong because lines of force cannot start from a negative charge.
(d) is wrong because lines of force cannot intersect each other.
(e) is wrong because electrostatic field lines cannot form closed loops.

When we start the motor, there is no back emf as the motor is at rest. So, a large current flows through the coil. As the motor rotates, the back emf increases and intake of current decreases.

The back emf in the coil opposes the change in the current as per Lenz's law.

Here, m = 10^{- 3} kg, q = 5 x 10^-6 C,
E = 2 x 10⁵ N/C , u = 20 m/s, v = 0
As it enters opposite to the field, so particle will retard.
Acceleration , 
\(a=\frac{qE}{m}\)
= \(\frac{5\times10^{-6}\times2\times10^5}{10{-3}}\)
= 10³ m/s²
Using,   v² = u² - 2as
0 = (20)² -2 x 1000 x s
or   s = \(\frac{400}{2000}\) 
= \(\frac{1}{5}\) 
= 0.2 m

As heat produced is the same, rms voltage of a.c. source = d.c. voltage = 110 V

Here, \(E={Q\over 4\pi\epsilon_or^2}=24NC^{-1}\)
\(V={Q\over 4\pi\epsilon_or}=12JC^{-1}\)
Dividing, we get \({V\over E}=r={12\over 24}=0.5m=AB\)
From V = \({Q\over 4\pi\epsilon_or};Q=4\pi\epsilon_or\times V\)
\(Q={1\over 9\times 10^9}\times0.5\times 12=0.667\times 10^{-9}C\)

\(Here, \ { E }_{ v }=60V, \ P=10W\)
\(R=\frac { { E }_{ v }^{ 2 } }{ P } =\frac { 60\times 60 }{ 10 } =360\Omega \)
\(Current \ through \ the \ lamp,\)
\({ I }_{ v }=\frac { { E }_{ v } }{ R } =\frac { 60 }{ 360 } =\frac { 1 }{ 6 } A\)
\(Let \ L \ be \ the \ inductance \ of \ the \ choke \ required.\)
\( \therefore \ \ Z=\frac { { E }_{ v } }{ { I }_{ v } } =\frac { 100 }{ 1/6 } =600\Omega \)
\( From \ \ { R }^{ 2 }+{ X }_{ L }^{ 2 }={ Z }^{ 2 }\)
\({ X }_{ L }=\sqrt { { Z }^{ 2 }-{ R }^{ 2 } } =\sqrt { { \left( 600 \right) }^{ 2 }-{ \left( 360 \right) }^{ 2 } } =480\Omega \)
\({ X }_{ L }=\omega L=2\pi vL=480\)
\(L=\frac { 480 }{ 2\pi v } =\frac { 480 }{ 2\times 3.14\times 60 } \)
\(L=1.274 \ H\)
\(To \ achive \ the \ same \ result \ resistance \ required\)
\({ R }^{ ' }=Z-R=600-360=240\Omega \)