1
MCQ (Single Correct Answer)

### JEE Main 2018 (Online) 16th April Morning Slot

In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery and a high resistance of 11 k$\Omega$ are used. The figure of merit of the galvanometer is 60 $\mu A/$division. In the absence of shunt resistance, the galvanometer produces a deflection of $\theta$ = 9 divisions when current flows in the circuit. The value of the shunt resistance that can cause the deflection of $\theta /2,$ is closest to :
A
500 $\Omega$
B
220 $\Omega$
C
55 $\Omega$
D
110 $\Omega$
2
MCQ (Single Correct Answer)

### JEE Main 2017 (Online) 9th April Morning Slot

In an experiment to determine the period of a simple pendulum of length 1 m, it is attached to different spherical bobs of radii r1 and r2 . The two spherical bobs have uniform mass distribution. If the relative difference in the periods, is found to be 5×10−4 s, the difference in radii, $\left| {} \right.$r1 $-$ r2 $\left| {} \right.$ is best given by :
A
1 cm
B
0.05 cm
C
0.5 cm
D
0.01 cm
3
MCQ (Single Correct Answer)

### JEE Main 2017 (Online) 9th April Morning Slot

In an experiment a convex lens of focal length 15 cm is placed coaxially on an optical bench in front of a convex mirror at a distance of 5 cm from it. It is found that an object and its image coincide, if the object is placed at a distance of 20 cm from the lens. The focal length of the convex mirror is :
A
27.5 cm
B
20.0 cm
C
25.0 cm
D
30.5 cm
4
MCQ (Single Correct Answer)

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

In an experiment a sphere of aluminium of mass 0.20 kg is heated upto 150oC. Immediately, it is put into water of volume 150 cc at 27oC kept in a calorimeter of water equivalent to 0.025 kg. Final temperature of the system is 40oC. The specific heat of aluminium is : (take 4.2 Joule = 1 calorie)
A
378 J/kg $-$oC
B
315 J/kg $-$oC
C
476 J/kg $-$oC
D
434 J/kg $-$oC

## Explanation

Let specific heat of aluminium = S,

As we know from principle of calorimetry,

Qgiven = Qused

$\therefore\,\,\,$ 0.2 $\times$ S $\times$ (150 $-$ 40) =
150 $\times$ 1 $\times$ (40 $-$ 27) + 25 $\times$ (40$-$27)

$\Rightarrow $$\,\,\, 0.2 \times S \times 110 = 150 \times 13 + 25 \times 13 \Rightarrow$$\,\,\,$ S = 434 J/kg - oC

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