## Three disc A, B and C having radii 2 m, 4 m and 6 m respectively are coated with carbon black on their outer surfaces…

Q: Three disc A, B and C having radii 2 m, 4 m and 6 m respectively are coated with carbon black on their outer surfaces. The wavelengths corresponding to maximum intensity are 3000 nm, 4000 nm and 5000 nm respectively. The power radiated by them are QA , QB and QC respectively.

(a) QA is maximum

(b) QB is maximum

(c) QC is maximum

(d) QA = QB = QC

Ans:(b)

Sol: As Q ∝ AT4 and λm T= constant.

$\displaystyle Q \propto \frac{A}{(\lambda_m)^4}$

$\displaystyle Q \propto \frac{r^2}{(\lambda_m)^4}$

$\displaystyle Q_A : Q_B : Q_C = \frac{2^2}{3^4} : \frac{4^2}{4^4} : \frac{6^2}{5^4}$

$\displaystyle Q_A : Q_B : Q_C = \frac{4}{81} : \frac{1}{16} : \frac{36}{625}$

i.e., QB is maximum.

## 2 kg of ice at – 20 °C is mixed with 5 kg of water at 20 °C in an insulating vessel a negligible heat capacity. Calculate the final mass of water…

Q: 2 kg of ice at – 20 °C is mixed with 5 kg of water at 20 °C in an insulating vessel a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water and ice are 1 kcal/kg/°C and 0.5 kcal/°C. While the latent heat of fusion of ice = 80 kcal/kg

(a) 7 kg

(b) 6 kg

(c) 4 kg

(d) 2 kg

Ans:(b)

Sol: Heat released by 5 kg of water when its temperature falls from 20 °C to 0°C is

Q1 = c m(ΔT)=103 × 5 × 20

= 0.2×105 cal.

This heat is used in raising the temp. of 2 kg of ice at – 20 °C to 0 °C and then melting it subsequently. Heat energy taken by 2 kg of ice at -20 °C to 0 °C and then melting it subsequently. Heat energy taken by 2 kg of ice at -20 °C in coming to 0 °C is

Q2 = c m (ΔT)

= 500 × 2 × 20

= 0.2 × 105 cal

The remaining heat

Q = Q1-Q2 =0.8×105 cal.

Mass of ice melted, m’= Q/L

=(0.8×105)/(80×103 ) = 1 kg.

Temperature of mixture will become 0 °C

Mass of water in it = 5 + 1 = 6 kg

Mass of ice left in it = 2 -1 = 1 kg.

## A planet is at an average distance d from the sun and its average surface temperature is T. Assume that…

Q:  A planet is at an average distance d from the sun and its average surface temperature is T. Assume that the planet receives energy only from the sun, and loss energy only through radiation from its surface. Neglect atmospheric effects. If T ∝ d -n , the value of n is

(a) 2

(b) 1

(c) 1/2

(d) 1/4

Ans: (c)

Sol: Let R = radius of planet

P = power radiated by the sun.

Energy received by planet per sec.

$\displaystyle = (\frac{P}{4\pi d^2})\times 4 \pi R^2$

Energy radiated by planet per sec.

=(4 π R2 )σ T4

For thermal equilibrium,

$\displaystyle 4 \pi R^2 (\sigma T^4)= (\frac{P}{4\pi d^2})\times 4 \pi R^2$

$\displaystyle T^4 \propto \frac{1}{d^2}$

$\displaystyle T \propto d^{-1/2}$

Hence , n = 1/2

## The rays of sun are focused on a piece of ice through a lens of diameter 5 cm, as a result of which 10 gram ice melts…

Q: The rays of sun are focused on a piece of ice through a lens of diameter 5 cm, as a result of which 10 gram ice melts in 10 min. The amount of heat received from sun, per unit area per min. is

(a) 4 cal/cm2 minute

(b) 40 cal/cm2-min

(c) 4 joule/m2 min

(d) 400 cal/cm2 min

Ans:(a)

Sol: Mass of ice melted/minute = 10/10 = 1 g

Quantity of heat used = 1 × 80 cal

Area of lens = π r2

= 3.14 (2.5)2

= 19.625 cm2

Amount of heat received/min/cm2 from sun

= 80/19.625

≈ 4 cal/cm2– min

## When a liquid in a glass vessel is heated, its apparent expansion is 10.30 × 10^-4/°C. When the same liquid is heated…

When a liquid in a glass vessel is heated, its apparent expansion is 10.30 × 10-4/°C. When the same liquid is heated in a metal vessel, its apparent expansion is 10.06×10-4/°C . If the coefficient of linear expansion of glass =9×10-6/°C, what is the coefficient of linear expansion of metal ?

(a) 51×10-6/°C

(b) 17× 10-6/°C

(c) 25×10-6/°C

(d) 43×10-6/°C

Ans: (b)

Sol : Here, γag = 10.30 × 10-4/°C

γam = 10.06 × 10-4/°C

αg=9×10-6/°C , αm = ?

Now, γr = γag + γg = γam + γm

γr = 10.30 × 10-4 + 3 × 9 × 10-6 = 10.06 × 10-4 + γm

γm = (10.30 + 0.27 – 10.06 ) 10-4

γm = 0.51 × 10-4

αm = γm/3 = 0.51 × 10-4/3

αm = 0.17 × 10-4/°C

αm = 17 × 10-6/°C