1

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

A 50 $\Omega$ resistance is connected to a battery of 5 V. A galvanometer of resistance 100 $\Omega$ is to be used as an ammeter to measure current through the resistance, for this a resistance rs is connected to the galvanometer. Which of the following connections should be employed if the measured current is within 1% of thecurrent without the ammeter in the circuit ?
A
rs = 0.5 $\Omega$ in parallel with the galvanometer
B
rs = 0.5 $\Omega$ in series with the galvanometer
C
rs = 1 $\Omega$ in series with galvanometer
D
rs =1 $\Omega$ in parallel with galvanometer

## Explanation

Current in the circuit without ammeter

I $=$ ${5 \over {50}} = 0.1$ A

$\therefore$   With ammeter current $=$ 0.1 $\times$ ${{99} \over {100}}$ = 0.099 A

With ammeter equivalent resistance,

Req = 50 + ${{100\,{r_s}} \over {100 + {r_s}}}$

$\therefore$   0.099 = ${5 \over {50 + {{100\,{r_s}} \over {100 + {r_s}}}}}$

$\Rightarrow$   50 + ${{100\,{r_s}} \over {100 + {r_s}}}$ = ${5 \over {0.099}}$

$\Rightarrow$   ${{100\,{r_s}} \over {100 + {r_s}}} = 0.5$

$\Rightarrow$   100 rs = 50 + 0.5rs

$\Rightarrow$   99.5 rs = 50

$\Rightarrow$   rs $=$ ${{50} \over {99.5}} = 0.5\,\Omega$
2

### 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
3

### 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.
4

### 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