1
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 10th April Morning Slot

The resistance of an electrical toaster has a temperature dependence given by R(T) = R0 [1 + $$\alpha $$(T − T0)] in its range of operation. At T0 = 300 K, R = 100 $$\Omega $$ and at T = 500 K, R = 120 $$\Omega $$. The toaster is connected to a voltage source at 200 V and its temperature is raised at a constant rate from 300 to 500 K in 30 s. The total work done in raising the temperature is :
A
400 $$\ln \,{{1.5} \over {1.3}}\,J$$
B
200 $$\ln \,{{2} \over {3}}\,J$$
C
60000 $$\ln \,{{6} \over {5}}\,J$$
D
300 J

Explanation

Given,

R = R6 [1 + $$\alpha $$ (T $$-$$ t0)]

120 = 100 [1 + $$\alpha $$ (500 $$-$$ 300)]

$$ \Rightarrow $$   200 $$\alpha $$ = $${1 \over 5}$$

$$ \Rightarrow $$   $$\alpha $$ = 10$$-$$3   oC$$-$$1

Temperature of the toaster raised from 300 K to 500 K in 30 s.

$$ \therefore $$   Increment in the temperature in time t,

$$\Delta $$T = $${{500 - 300} \over {30}}t$$

= $${{200} \over 3}t$$

= $${{20} \over 3}t$$

Total work done in raising the temperature

= $$\int\limits_0^t {{{{V^2}} \over {R\left( t \right)}}\,dt} $$

= $$\int\limits_0^t {{{{V^2}} \over {{R_0}\left( {1 + \alpha \Delta t} \right)}}} \,dt$$

= $$\int\limits_0^{30} {{{{{\left( {200} \right)}^2}} \over {100\left( {1 + {{10}^{ - 3}} \times {{20} \over 3}t} \right)}}} \,dt$$

= $${{40000} \over {100}}\int\limits_0^{30} {{{dt} \over {\left( {1 + {t \over {150}}} \right)}}} $$

$$400 \times 150\left[ {\ln \left( {1 + {t \over {150}}} \right)} \right]_0^{30}$$

$$ = 60000\left[ {\ln \left( {1 + {{30} \over {150}}} \right) - \ln 1} \right]$$

$$ = 60000\ln \left( {{6 \over 5}} \right)\,J$$
2
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be:
A
$$CE{{{r_1}} \over {({r_1} + r)}}$$
B
CE
C
$$CE{{{r_1}} \over {({r_2} + r)}}$$
D
$$CE{{{r_2}} \over {(r + {r_2})}}$$

Explanation

In steady state, flow fo current through capacitor will be zero.

Current through the circuit,

i = $${E \over {r + {r_2}}}$$

Potential difference through capacitor

Vc = $${Q \over C}$$ = E - ir = E - $$\left( {{E \over {r + {r_2}}}} \right)r$$

$$ \therefore $$ Q = $$CE{{{r_2}} \over {(r + {r_2})}}$$
3
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

Which of the following statements is false?
A
Kirchhoff’s second law represents energy conservation.
B
Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude.
C
In a balanced wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed.
D
A rheostat can be used as a potential divider.

Explanation

There is no change in null point, if the cell and the galvanometer are exchanged in a balanced wheatstone bridge.
4
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

In the given circuit, the current in each resistance is:
A
0 A
B
1 A
C
0.25 A
D
0.5 A

Explanation

The potential difference in each loop is zero.

$$ \therefore $$ No current will flow or current in each resistance is Zero.

Questions Asked from Current Electricity

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