1

### JEE Main 2016 (Online) 9th April Morning Slot

To know the resistance G of a galvanometer by half deflection method, a battery of emf VE and resistance R is used to deflect the galvanometer by angle $\theta$. If a shunt of resistance S is needed to get half deflection then G, R and S are related by the equation :
A
2S (R + G) = RG
B
S (R + G) = RG
C
2S = G
D
2G = S

## Explanation

When only galvanometer G is present with the resistance R,

Here IG = ${{{V_E}} \over {R + G}}$

When shunt of resistance S is connected parallel to galvanometer,

Here I = ${{{V_E}} \over {R + {{GS} \over {G + S}}}}$

As deflection is half, here current through galvanometer,

IG' = ${{{{\rm I}_G}} \over 2}$

As both Galvanometer and shunt are parallel then potential are parallel then potential difference same.

$\therefore$   IG' (G) = (I $-$ IG')S

$\Rightarrow$   I'G (G + S) = IS

$\Rightarrow$   ${{{{\rm I}_G}} \over 2}$ = ${{{\rm I}S} \over {G + S}}$

$\Rightarrow$   ${{{V_E}} \over {2\left( {R + G} \right)}}$ = ${{{V_E}} \over {R + {{GS} \over {G + S}}}}$ $\times$ ${S \over {\left( {G + S} \right)}}$

$\Rightarrow$   ${1 \over {2\left( {R + G} \right)}}$ = ${{G + S} \over {R(G + S) + GS}}$ $\times$ ${S \over {\left( {G + S} \right)}}$

$\Rightarrow$   RG + RS + GS = 2RS + 2GS

$\Rightarrow$   RG = RS + GS

$\Rightarrow$    S(R + G) = RG
2

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

A fighter plane of length 20 m, wing span (distance from tip of one wing to the tip of the other wing) of 15 m and height 5 m is flying towards east over Delhi. Its speed is 240 ms−1. The earth’s magnetic field over Delhi is 5 $\times$10−5 T with the declination angle ~ 08 and dip of $\theta$ such that sin $\theta$ = 2 3 . If the voltage developed is VB between the lower and upper side of the plane and VW between the tips of the wings then VB and VW are close to :
A
VB = 45 mV; VW = 120 mV with right side of pilot at higher voltage
B
VB = 45 mV; VW = 120 mV with left side of pilot at higher voltage
C
VB = 40 mV; VW = 135 mV with right side of pilot at high voltage
D
VB = 40 mV; VW = 135 mV with left side of pilot at higher voltage

## Explanation

VB = vhBcos$\theta$

= 240 $\times$ 5 $\times$ 5 $\times$ 10$-$5$\times$ ${{\sqrt 5 } \over 3}$

= 44.7 $\times$ 10$-$3 V

= 45 mV

Vw = $l$vB sin$\theta$

= 15 $\times$ 240 $\times$ 5 $\times$ 10$-$5 $\times$ ${2 \over 3}$

= 1200 $\times$ 10$-$4 V

= 120 mV

From right hand rule, the charge moves to the left side of the pilot.
3

### 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
4

### 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 $\Omega$