A clock pendulum made of invar has a period o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.Taking the logarithmic derivative, we get $\frac{dT}{T} = \frac{1}

Vedclass · Accepted Answer

$2.25 	imes 10^{-6} \, s$

Vedclass · Answer

(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.Taking the logarithmic derivative, we get $\frac{dT}{T} = \frac{1}{2} \frac{dl}{l}$.Since the change in length is given by $\frac{dl}{l} = \alpha \Delta 	heta$, where $\Delta 	heta$ is the change in temperature, we have $\frac{dT}{T} = \frac{1}{2} \alpha \Delta 	heta$.Given $\alpha = 9 	imes 10^{-7} /{ }^{\circ} C$, $\Delta 	heta = 30^{\circ} C - 20^{\circ} C = 10^{\circ} C$, and $T = 0.5 \, s$.The change in time period (loss in time per oscillation) is $dT = T 	imes \frac{1}{2} \alpha \Delta 	heta$.Substituting the values: $dT = 0.5 	imes \frac{1}{2} 	imes 9 	imes 10^{-7} 	imes 10$.$dT = 0.5 	imes 4.5 	imes 10^{-6} = 2.25 	imes 10^{-6} \, s$.

$A$ clock pendulum made of invar has a period of $0.5 \, s$ at $20^{\circ} C$. If the clock is used in a climate where the temperature averages to $30^{\circ} C$, how much time does the clock lose in each oscillation? (For invar, $\alpha = 9 \times 10^{-7} /{ }^{\circ} C$, $g = \text{constant}$)

Similar Questions

The limiting friction between two bodies in contact is independent of

Which among the following reactions will take place spontaneously?

Water falls from a height of $60\,m$ at the rate of $15\,kg/s$ to operate a turbine. The losses due to frictional forces are $10\%$ of energy. How much power is generated by the turbine in $kW$ $(g = 10\,m/s^2)$?

In the reaction,$C_2H_5OC_2H_5 + CO \xrightarrow[150^{\circ}C, 500 \text{ atm pressure}]{BF_3} X$. What is $X$?

$(b \times c) \times (c \times a) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module