Question 1: Palette 1/7

DPT MCQ

Concept: Thermal expansion in solids (Linear expansion)

A steel ruler exactly 20 cm long is graduated to give correct measurements at $20^\circ C$. What will be the actual length of the ruler when it is used in the desert at a temperature of $40^\circ C$? (Take $\alpha_{steel} = 1.2 \times 10^{-5} / ^\circ C$)

A

$20.0024$ cm

B

$20.0048$ cm

C

$19.9952$ cm

D

$20.048$ cm

Question 2: Palette 2/7

DPT MCQ

Concept: Linear expansion and condition for constant difference in lengths

The length of a steel rod is 5 cm longer than that of a brass rod. If this difference in their lengths is to remain same at all temperatures, then find the length of the brass rod. (Coefficients of linear expansion for steel and brass are $12 \times 10^{-6} / ^\circ C$ and $18 \times 10^{-6} / ^\circ C$, respectively.)

A

15 cm

B

5 cm

C

10 cm

D

20 cm

Question 3: Palette 3/7

DPT MCQ

Concept: Superficial (areal) expansion of solids

A metal ball having a diameter of 0.4 m is heated from 273K to 360 K. If the coefficient of areal expansion of the material of the ball is $0.000034 K^{-1}$, then determine the increase in surface area of the ball.

A

$1.486 \times 10^{-3} m^2$

B

$2.486 \times 10^{-3} m^2$

C

$0.5024 \times 10^{-3} m^2$

D

$3.486 \times 10^{-3} m^2$

Question 4: Palette 4/7

DPT MCQ

Concept: Volume expansion and its relation to linear expansion

On heating a glass block of $10000 cm^3$ from $25^\circ C$ to $40^\circ C$, its volume increases by $4 cm^3$. Determine the coefficient of linear expansion of glass.

A

$8.89 \times 10^{-6} / ^\circ C$

B

$26.67 \times 10^{-6} / ^\circ C$

C

$13.33 \times 10^{-6} / ^\circ C$

D

$4.44 \times 10^{-6} / ^\circ C$

Question 5: Palette 5/7

DPT MCQ

Concept: Volume expansion of liquids and its application in a thermometer

The volume of mercury in the bulb of a thermometer is $10^{-6} m^3$. The area of cross-section of the capillary tube is $2 \times 10^{-7} m^2$. If the temperature is raised by $100^\circ C$, then find the increase in the length of the mercury column. (Take $\gamma_{Hg} = 18 \times 10^{-5} / ^\circ C$)

A

$4.5 cm$

B

$9 cm$

C

$18 cm$

D

$0.9 cm$

Question 6: Palette 6/7

DPT MCQ

Concept: When the temperature of a pendulum clock increases, its length increases, causing its time period to increase. The clock runs slow and loses time. The time lost or gained in a day is given by the formula $\Delta t = \frac{1}{2} \alpha \Delta \theta \times 86400$, where $\alpha$ is the coefficient of linear expansion, $\Delta \theta$ is the change in temperature, and 86400 is the number of seconds in a day.

A clock which keeps correct time at 20°C is subjected to 40°C. If the coefficient of linear expansion of the pendulum is $12 \times 10^{-6} /^\circ C$, determine the loss in time per day.

A

10.4 s

B

12.4 s

C

8.2 s

D

15.6 s

Question 7: Palette 7/7

DPT MCQ

Concept: When the temperature of a pendulum clock increases, its length increases, causing its time period to increase. The clock runs slow and loses time. The change in time period is given by the formula $\Delta T = \frac{1}{2} \alpha \Delta \theta T$. The time lost or gained is given by the formula $\Delta t = \frac{\Delta T}{T'} \times t$, where $T'$ is the new time period and $t$ is the total time elapsed.

A second's pendulum clock has a steel wire. The clock is calibrated at $20^\circ C$. How much time does the clock loss or gain in one week when the temperature is increased to $30^\circ C$? (Take $\alpha_{steel} = 1.2 \times 10^{-5} /^\circ C$)

A

42.15 s

B

28.50 s

C

36.28 s

D

50.00 s