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A sphere, a cube and a thin circular plate are allmade of the same material, have the same mass and are initially heated to a temperature of 200°C.Arrange then in the ascending order of rate of cooling.

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Explanation

$ { d \theta \over dt } \alpha A $now, for a given mass,sphere has the least surface area and circular plate will have the maximum surface area. Hence the sphere will cool the slowest and the disc the fastest.

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A tube, closed at one end and containing air, produces, when excited,the fundamental note of frequency 512 Hz. If the tube is opened at both ends what is the fundamental frequency that can be excited (in Hz) ?

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Explanation

$ f_0 = { v \over 4 l } \Rightarrow f_0 = { v \over 2l } = 2 f_c $

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An open pipe is suddenly closed at one end with the result that the frequency of third harmoni of the closed pipe is found to be higher by 100 Hz than the fundamental frequency of the open pipe. What is the fundamental frequency of the open pipe ?

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Explanation

Fundamental frequency of open pipe is $ f_1 = { \upsilon \over 2 l } $ and Frequency of third harmonic of closed pipe will be $ f_2 = 3 \left( \upsilon \over 4 l \right) $ Given that $ f_2 - f_1 = 100 $

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Two vibrating strings of the same material but of lengths Land 2L have radii 2r and r respectively. They are stretched under the same tension. Both the string vibrate in their fundamental modes. The one of length L with frequency $\upsilon_1$ and the other with frequency $ \upsilon_2$ . What is the ratio $ { \upsilon_1 \over \upsilon_2} $ ?

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Explanation

Fundamental frequency is given by $ v = { 1 \over 2l } \sqrt { T \over \mu } $ (with both the end fixed) $ \therefore $Fundamentalfrequency $ \upsilon \alpha { 1 \over l \sqrt \mu } $ (for same tension in both strings) where $ \mu $ = mass per unit length of wire $ = \rho . A $ $ = \rho ( \pi r^2 ) or \sqrt \mu \alpha r $ $ \therefore \upsilon \alpha { 1 \over rl } $ $ \therefore { \upsilon_1 \over \upsilon_2 } = \left( { r_2 \over r_1 } \right) \left( { l_2 \over l_1 } \right) = \left( { r \over 2r } \right) \left( { 2L \over L } \right) = 1 $

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A closed organ pipe of length L and an open organ pipe contain gases of densities $n_1$ and $n_2$ respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibratingin their first overtone with same frequency. What is the length of the open organ pipe ?

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Explanation

$ f_c = f_0 $( both first overtone ) $ \Rightarrow 3 \left( {\upsilon_c \over 4L } \right) = 2 \left({ \upsilon_0 \over 2 l_0 } \right) $ $ \therefore l_0 = {4 \over 3 } \left( {\upsilon_0 \over \upsilon_c} \right) L = { 4 \over 3 } \sqrt { \rho_1 \over \rho_2 } L \;as\; \upsilon \;\alpha { 1 \over \sqrt p } $

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An open pipe is in resonance in 2nd harmonic with frequency $f_1$ . Now one end of the tube is closed and frequency is increased to $f_2$ such that the resonance again occurs in nth harmonic. Choose the correct option.

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Explanation

$ f_1 = { \upsilon \over l} $ (2nd harmonic of open pipe) $ f_2 = n \left( {\upsilon \over 4l} \right ) $ (nth harmonic of closed pipe) Here, n is odd and $f_2 \gt f_1 $ it is possible when n = 5 because with n = 5 $ f_2 = {5 \over 4 } \left( {\upsilon \over l } \right ) = {5 \over 4 } f_1 $

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A tuning fork of 512 Hz is used to produce resonance ina resonance tube experiment. The level of water at first resonance is 30.7 cm and at second resonance is 63.2 cm. What is the error in calculating velocity of sound ?Asume the speed of sound 330 m/s.

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Explanation

$ { \lambda \over 2 } = (63.2 - 30.7 ) cm or \lambda = 0.65 m $ $ \therefore ,speed of sound observed, v_0 = f \lambda $

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An air column in a pipe, which is closed at one end will be in resonance with a ribrating tuning fork of frequency 264 Hz, What is the length of the column if it is in cm ? (speed of sound in air = 330 m/s)

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Explanation

For closed organ pipe $ f = n \left( {\upsilon \over 4 l } \right) $ where n = 1,3,5 ..... $ l = { nv \over 4f } $

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In a resonance tube experiment, the first resonance is obtained for 10 cm of air column and the second for 32 cm.The end correction for this apparatus is equal to

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Explanation

End correction = $ { l_2 -3 l_1 \over 2 } $

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Amount of heat required to raise the temperature of a body through 1k is called its

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Explanation

$ \theta = m .C .\triangle \theta, if \triangle \theta = 1k $

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