Question

A thin conducting rod MN of mass 20 gm, length 25 cm, and resistance 10 Ω are held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is a uniform magnetic field 𝐵₀ = 4 T directed perpendicular to the plane of the rod-rail arrangement. The rod is released from rest at time 𝑡 = 0 and it moves down along the rails. Assume air drag is negligible. Match each quantity in List-I with an appropriate value from List-II, and choose the correct option. 
[Given: The acceleration due to gravity 𝑔 = 10 m s⁻² and 𝑒⁻¹ = 0.4]

List I		List II
(P)	At 𝑡 = 0.2 s, the magnitude of the induced emf in Volt	(1)	0.07
(Q)	At 𝑡 = 0.2 s, the magnitude of the magnetic force in Newton	(2)	0.14
(R)	At 𝑡 = 0.2 s, the power dissipated as heat in Watt	(3)	1.20
(S)	The magnitude of terminal velocity of the rod in m s−1	(4)	0.12
		(5)	2.00

 

• A. P$\rightarrow$5, Q$\rightarrow$2, R$\rightarrow$3, S$\rightarrow$1
• B. P$\rightarrow$3, Q$\rightarrow$1, R$\rightarrow$4, S$\rightarrow$5
• C. P$\rightarrow$4, Q$\rightarrow$3, R$\rightarrow$1, S$\rightarrow$2
• D. P$\rightarrow$3, Q$\rightarrow$4, R$\rightarrow$2, S$\rightarrow$5

Answer 1

The Correct Option is D

Induced emf = $\varepsilon=Blv$
Induced current = $i=\frac{\varepsilon}{R}=\frac{Blv}{R}$
$\Rightarrow mg-ilB=ma$
$\Rightarrow mg-\frac{B^2l^2v}{R}=m\frac{dv}{dt}$
$\Rightarrow \frac{dv}{mg-\frac{B^2l^2v}{R}}=\frac{dt}{m}$
$\Rightarrow \frac{mg-\frac{B^2l^2v}{R}}{mg}=e^{\frac{-{B^2l^2}}{mR}t}$
$\Rightarrow v=2[1-e^{-5t}]$
At t = 0.2s, v = $2[1-\frac{1}{e}]$
$\Rightarrow \varepsilon=Bl\times 2[1-\frac{1}{e}]$=1.2 volt.
Magnetic force = $ilB$ = 0.12N
The power dissipated = 0.144W
Terminal velocity = 2 m/s