1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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 $$\Omega $$

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