A thin conducting rod MN of mass 20 text{ g}, | vedclass

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(D) Given: $m = 20 	imes 10^{-3} 	ext{ kg}$,$\ell = 0.25 	ext{ m}$,$R = 10 	ext{ }\Omega$,$B = 4 	ext{ T}$,$g = 10 	ext{ m s}^{-2}$.The equation

Vedclass · Accepted Answer

$P \rightarrow 3, Q \rightarrow 4, R \rightarrow 2, S \rightarrow 5$

Vedclass · Answer

(D) Given: $m = 20 	imes 10^{-3} 	ext{ kg}$,$\ell = 0.25 	ext{ m}$,$R = 10 	ext{ }\Omega$,$B = 4 	ext{ T}$,$g = 10 	ext{ m s}^{-2}$.The equation of motion is $mg - Bi\ell = m \frac{dv}{dt}$. Since $i = \frac{B\ell v}{R}$,we have $mg - \frac{B^2\ell^2 v}{R} = m \frac{dv}{dt}$.Rearranging gives $\frac{dv}{dt} = \frac{B^2\ell^2}{mR} (v_T - v)$,where $v_T = \frac{mgR}{B^2\ell^2}$ is the terminal velocity.$v_T = \frac{20 	imes 10^{-3} 	imes 10 	imes 10}{4^2 	imes (0.25)^2} = \frac{2}{16 	imes 0.0625} = \frac{2}{1} = 2 	ext{ m s}^{-1}$.The time constant $	au = \frac{mR}{B^2\ell^2} = \frac{20 	imes 10^{-3} 	imes 10}{1} = 0.2 	ext{ s}$.The velocity at time $t$ is $v(t) = v_T(1 - e^{-t/	au}) = 2(1 - e^{-t/0.2})$.At $t = 0.2 	ext{ s}$,$v = 2(1 - e^{-1}) = 2(1 - 0.4) = 1.2 	ext{ m s}^{-1}$.$(P)$ Induced emf $E = Bv\ell = 4 	imes 1.2 	imes 0.25 = 1.2 	ext{ V}$. (Matches $3$)$(Q)$ Magnetic force $F_m = Bi\ell = B(\frac{Bv\ell}{R})\ell = \frac{B^2\ell^2 v}{R} = \frac{16 	imes 0.0625 	imes 1.2}{10} = 0.12 	ext{ N}$. (Matches $4$)$(R)$ Power $P = i^2R = (\frac{E}{R})^2 R = \frac{E^2}{R} = \frac{(1.2)^2}{10} = 0.144 	ext{ W}$. (Matches $2$)$(S)$ Terminal velocity $v_T = 2.00 	ext{ m s}^{-1}$. (Matches $5$)Thus,$P ightarrow 3, Q ightarrow 4, R ightarrow 2, S ightarrow 5$. The correct option is $(D)$.

List-$I$	List-$II$
$(P)$ At $t = 0.2 \text{ s}$,the magnitude of the induced emf in Volt	$(1)$ $0.07$
$(Q)$ At $t = 0.2 \text{ s}$,the magnitude of the magnetic force in Newton	$(2)$ $0.144$
$(R)$ At $t = 0.2 \text{ s}$,the power dissipated as heat in Watt	$(3)$ $1.20$
$(S)$ The magnitude of terminal velocity of the rod in $\text{m s}^{-1}$	$(4)$ $0.12$
	$(5)$ $2.00$

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