An extended object is placed at a point O , 10 cm in front of a convex lens L1 and a concave lens L2 is placed ….

Q: An extended object is placed at a point O , 10 cm in front of a convex lens L1 and a concave lens L2 is placed 10 cm behind it , as shown in the figure . The radii of curvature of all the curved surface in both the lenses are 20 cm .The refractive index of both the lenses is 1.5 . The total magnification of this lens system is

Numerical

(a) 0.4

(b) 0.8

(c) 1.3

(d) 1.6

Click to See Solution :
Ans: (b)
Sol: Focal length of Convex lens f1

$\displaystyle \frac{1}{f_1} = (\mu -1)[\frac{1}{f_1} – \frac{1}{f_2}]$

$\displaystyle \frac{1}{f_1} = (1.5 -1)[\frac{1}{20} – \frac{1}{-20}]$

f1 = 20 cm

Focal length of Concave lens f2

$\displaystyle \frac{1}{f_2} = (\mu -1)[\frac{1}{f_1} – \frac{1}{f_2}]$

$\displaystyle \frac{1}{f_2} = (1.5 -1)[\frac{1}{-20} – \frac{1}{20}]$

f2 = -20 cm

For Lens 1 :

$\displaystyle \frac{1}{v} – \frac{1}{u} = \frac{1}{f}$

v = – 20 cm
$\displaystyle m_1 = \frac{v}{u} = \frac{-20}{-10} = 2$

For Lens 2 :

u = -30 cm , f = -20 cm

$\displaystyle \frac{1}{v} – \frac{1}{u} = \frac{1}{f}$

v = -12 cm

$\displaystyle m_2 = \frac{v}{u} = \frac{-12}{-30} = \frac{2}{5}$

Net magnification m = m1 x m2

m = 2 x (2/5) = 4/5 =0.8

 

An Ideal gas undergoes a four step cycle as shown in P-V diagram below . During this cycle …

Q: An Ideal gas undergoes a four step cycle as shown in P-V diagram below . During this cycle , heat is absorbed by the gas in

IIT

(a) Steps 1 and 2

(b) Steps 1 and 3

(c) Steps 1 and 4

(d) Steps 2 and 4

Click to See Solution :
Ans: (c)

Sol:Process 1:

P = constant , Volume increases and temperature also increases

Q = W + ΔU

W = +ve , ΔU = positive

Heat is positive and supplied to the gas

Process 2 :
V = constant , Pressure decreases

T ∝ P , (V = constant )

Temperature decreases , W = 0
ΔT is negative and ΔU = f/2 nRΔT

ΔU is also negative

Q = ΔU + W

Heat is negative and rejected by gas

Process 3 :
P = constant , Volume decreases

Temperature also decreases

W = PΔV = negative

ΔU = f/2 (nRΔT) = negative

Heat is negative and rejected by gas

Process 4 :
V = constant , Pressure increases

W = 0
PV = n RT => temperature increases

ΔU = f/2 (nRΔT) is positive

ΔQ = ΔU + W = Positive

 

The smallest division on the main scale of a Vernier calipers is 0.1 cm . Ten division of the vernier scale correspond to nine division of main scale ….

Q: The smallest division on the main scale of a Vernier calipers is 0.1 cm . Ten division of the vernier scale correspond to nine division of main scale . The figure below on the left shows the reading of this calipers with no gap between its two jaws . The figure on the right shows the reading with a solid sphere . The figure on the right shows the reading with a solid sphere held between the jaws . The correct diameter of the sphere is

IIT

(a)3.07 cm (b) 3.11 cm (iii) 3.15 cm (iv) 3.17 cm

Click to See Solution :
Ans: (c)

Sol: Least Count $= (1 – \frac{9}{10})0.1 = 0.01 cm $

Zero Error = -0.1 + 0.06 = -0.04 cm

Final Reading = 3.1 + 0.01 x 1 = 3.11 cm

So , Correct measurement = 3.11 + 0.04 = 3.15 cm

 

A container with 1 kg of water in it is kept in sunlight , which causes the water to get warmer than the surroundings .

Q: A container with 1 kg of water in it is kept in sunlight , which causes the water to get warmer than the surroundings . The average energy per unit time per unit area received due to the sunlight is 700 W/m2 and it is absorbed by the water over an effective area of 0.05 m2 . Assuming that the heat loss from the water to the surroundings is governed by Newton’s Law of cooling , the difference (in °C ) in the temperature of water and the surroundings after a long time will be ……….. (Ignore effect of the container and take constant for Newton’s Law of cooling = 0.001 s-1 , Heat capacity of water = 4200 JKg-1 K-1 )

Click to See Solution :
Ans: (8.33)

Sol: $\displaystyle \frac{dQ}{dt} = e \sigma A(T^4 – T_o^2)$

For small temperature change ,

$\displaystyle \frac{dQ}{dt} = e \sigma A T^3 \Delta T $ …(i)

$\displaystyle \frac{mC dT}{dt} = e \sigma A T^3 \Delta T $

$\displaystyle \frac{dT}{dt} = \frac{e \sigma A T^3}{m C} \Delta T $

$\displaystyle \frac{e \sigma A T^3}{m C} $ -> Constant for Newton’s Law of cooling

$\displaystyle \frac{e \sigma A T^3}{m C} = 0.001 $

$\displaystyle e \sigma A T^3 = m c \times 0.001 = 1 \times 4200 \times 0.001 $

$\displaystyle e \sigma A T^3 = 4.2 $ …(ii)

$\displaystyle \frac{dQ}{dt} = 700 \times 0.05 = 35 W $ ….(iii)

Putting the value of Eqs (ii) & (iii) in Eq.(i)

35 = 4.2 ΔT

ΔT = 35/4.2 = 8.33