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.

### EXAM MAP

#### Joint Entrance Examination

JEE Advanced JEE Main

#### Graduate Aptitude Test in Engineering

GATE CE GATE ECE GATE ME GATE IN GATE EE GATE CSE GATE PI

NEET