1

### 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 $\Omega$
2

### 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 $\Omega$
3

### 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
4

### JEE Main 2017 (Online) 8th April Morning Slot

In a certain region static electric and magnetic fields exist. The magnetic field is given by $\overrightarrow B = {B_0}\left( {\widehat i + 2\widehat j - 4\widehat k} \right)$ . If a test charge moving with a velocity $\overrightarrow \upsilon = {\upsilon _0}\left( {3\widehat i - \widehat j + 2\widehat k} \right)$ experiences no force in that region, then the electric field in the region, in SI units, is :
A
$\overrightarrow E = - {\upsilon _0}\,{B_0}\left( {3\widehat i - 2\widehat j - 4\widehat k} \right)$
B
$\overrightarrow E = - {\upsilon _0}\,{B_0}\left( {\widehat i + \widehat j + 7\widehat k} \right)$
C
$\overrightarrow E = {\upsilon _0}\,{B_0}\left( {14\widehat j + 7\widehat k} \right)$
D
$\overrightarrow E = - {\upsilon _0}\,{B_0}\left( {14\widehat j + 7\widehat k} \right)$

## Explanation

Here test charge experience no net force, So, sum of electric and magnetic field is zero.

$\therefore\,\,\,$ Fe + Fm = 0

$\therefore\,\,\,$ Fe = $-$ q ($\overrightarrow v$ $\times$ $\overrightarrow B)$

= $-$ qB0 $\upsilon$0 [(3$\widehat i$ $-$ $\widehat j$ + 2$\widehat k$) $\times$ ($\widehat i$ + 2$\widehat j$ $-$ 4$\widehat k$)]

= $-$ q$\upsilon$0 B0 (14$\widehat j$ + 7$\widehat k$)

Electric field produced by the charge q,

$\overrightarrow E$ = ${{\overrightarrow {{F_e}} } \over q}$

= ${{ - q{\upsilon _0}{B_0}\left( {14\widehat j + 7\widehat k} \right)} \over q}$

= $-$ $\upsilon$0 B0 (14 ${\widehat j}$ + 7 ${\widehat k}$)