1

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

To find the focal length of a convex mirror, a student records the following data :

Object
Pin
Convex
Lens
Convex
Mirror
Image
Pin
22.2 cm 32.2 cm 45.8 cm 71.2 cm

The focal length of the convex lens is f1 and that of mirror is f2. Then taking index correction to be negligibly small, f1 and f2 are close to :
A
f1 = 12.7 cm    f2 = 7.8 cm
B
f1 = 7.8 cm    f2 = 12.7 cm
C
f1 = 7.8 cm    f2 = 25.4 cm
D
f1 = 15.6 cm    f2 = 25.4 cm

## Explanation

For lens :

u1 = $-$ (32.2 $-$ 22.2) cm

= $-$ 10 cm

v1 = (71.2 $-$ 32.2) cm

= 39 cm

$\therefore$   ${1 \over {{f_1}}}$ = ${1 \over {{v_1}}} - {1 \over {{u_1}}}$

= ${1 \over {39}}$ + ${1 \over {10}}$

= ${{49} \over {390}}$

$\therefore$   f1 = 7.8 cm

For mirror :

R = (71.2 $-$ 45.8) cm

= 25.4 cm

$\therefore$   f2 = ${R \over 2}$ = ${{25.4} \over 2}$ = 12.7 cm
2

### 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
3

### 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

### 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