1

### JEE Main 2016 (Online) 10th April Morning Slot

Consider a thin metallic sheet perpendicular to the plane of the paper moving with speed ‘v’ in a uniform magnetic field B going into the plane of the paper (See figure). If charge densities $\sigma$1 and $\sigma$2 are induced on the left and right surfaces, respectively, of the sheet then (ignore fringe effects) :

A
$\sigma$1 = $\in$0 $\upsilon$ B, $\sigma$2 = $-$ $\in$0 $\upsilon$ B
B
$\sigma$1 = ${{{ \in _0}\upsilon \,B} \over 2},$ $\sigma$2 = ${{ - { \in _0}\,\upsilon B} \over 2}$
C
$\sigma$1 = $\sigma$2 = ${ \in _0}\,\upsilon B$
D
$\sigma$1 = ${{ - { \in _0}\upsilon B} \over 2},$ $\sigma$2 = ${{ { \in _0}\upsilon B} \over 2},$

## Explanation

Magnetic force on electron

$\overrightarrow F$ = $-$ e $\left( {\overrightarrow V \times \overrightarrow B } \right)$

F = eVB   [As ${\overrightarrow V }$ and ${\overrightarrow B }$ are perpendicular]

Also,   F = e E

and  E = ${\sigma \over {{\varepsilon _0}}}$

$\therefore$   eVB = e $\times$ ${\sigma \over {{\varepsilon _0}}}$

$\Rightarrow$   $\sigma$ = ${{\varepsilon _0}}$VB = $\sigma$1

as   $\sigma$1 = $-$ $\sigma$2

$\therefore$   $\sigma$2 = $-$ ${{\varepsilon _0}}$VB
2

### JEE Main 2016 (Online) 10th April Morning Slot

A galvanometer has a 50 division scale. Battery has no internal resistance. It is found that there is deflection of 40 divisions when R = 2400 $\Omega$. Deflection becomes 20 divisions when resistance taken from resistance box is 4900 $\Omega$. Then we can conclude :

A
Resistance of galvanometer is 200 $\Omega$
B
Full scale deflection current is 2 mA.
C
Current sensitivity of galvanometer is 20 $\mu$A/division.
D
Resistance required on R.B. for a deflection of 10 divisions is 9800 $\Omega$.

## Explanation

Let full scale deflection of current = I

In case 1, when R = 2400 $\Omega$ and deflection of 40 divisions present.

$\therefore$   ${{40} \over {50}}{\rm I} = {V \over {G + R}}$

$\Rightarrow$   ${4 \over 5}{\rm I}$ = ${2 \over {G + 2400}}$          . . .(1)

Incase 2, when R = 4900 $\Omega$ and deflection of 20 divisions present

$\therefore$   ${{20} \over {50}}{\rm I} = {V \over {G + R}}$

$\Rightarrow$   ${2 \over 5}{\rm I}$ = ${2 \over {G + 4900}}$          . . .(2)

From (1) and (2) we get,

${4 \over 2} = {{G + 4900} \over {G + 2400}}$

$\Rightarrow$   2G + 4800 $=$ G + 4900

$\Rightarrow$   G $=$ 100 $\Omega$

Putting value of G in equation (1), we get,

${4 \over 5}{\rm I} = {2 \over {100 + 2400}}$

$\Rightarrow$   ${\rm I} = 1$ mA

Current sensitivity $=$ ${{\rm I} \over {number\,\,of\,\,divisions}}$

$=$ ${1 \over {50}}$

$=$ 0.02 mA / division

$=$ 20 $\mu$A / division

Resistance required for deflection of 10 divisions

${{10} \over {50}}{\rm I} = {V \over {G + R}}$

$\Rightarrow$   ${1 \over 5} \times 1 \times {10^{ - 3}} = {2 \over {100 + R}}$

$\Rightarrow$   R $=$ 9900 $\Omega$
3

### JEE Main 2017 (Offline)

When a current of 5 mA is passed through a galvanometer having a coil of resistance 15$\Omega$, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0 – 10V is:
A
4.005 × 103 $\Omega$
B
1.985 × 103 $\Omega$
C
2.535 × 103 $\Omega$
D
2.045 × 103 $\Omega$

## Explanation

Given : Current through the galvanometer,

ig = 5 × 10–3 A

Galvanometer resistance, G = 15 $\Omega$

Let resistance R to be put in series with the galvanometer to convert it into a voltmeter.

V = ig (R + G)

10 = 5 × 10–3 (R + 15)

$\therefore$ R = 2000 – 15 = 1985

= 1.985 × 103 $\Omega$
4

### JEE Main 2017 (Offline)

A magnetic needle of magnetic moment 6.7 $\times$ 10-2 A m2 and moment of inertia 7.5 $\times$ 10-6 kg m2 is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is:
A
8.76 s
B
6.65 s
C
8.89 s
D
6.98 s

## Explanation

Given : Magnetic moment, M = 6.7 × 10–2 Am2

Magnetic field, B = 0.01 T

Moment of inertia, I = 7.5 × 10–6 Kgm2

Using, T = $2\pi \sqrt {{I \over {MB}}}$

= $2\pi \sqrt {{{7.5 \times {{10}^{ - 6}}} \over {6.7 \times {{10}^{ - 2}} \times 0.01}}}$

= ${{2\pi } \over {10}} \times 1.06$ s

Time taken for 10 complete oscillations

t = 10T = 2$\pi$ × 1.06

= 6.6568 $\simeq$ 6.65 s