A particle is undergoing SHM having total mechanical energy equal to $8text{ J}$. At an instant its kinetic energy is found to be $10text{ J}$, then its potential energy at that instant is
The total mechanical energy $E$ is the sum of kinetic and potential energy: $E = K + U$. Thus, $8 = 10 + U implies U = -2text{ J}$.
Time period of a second’s pendulum is $2text{ s}$, the approximate length of its string is equal to ($g = 10text{ m/s}^2$)
Using the formula $T = 2pi sqrt{frac{l}{g}}$, we substitute $T = 2text{ s}$ and $g = 10text{ m/s}^2$: $2 = 2pi sqrt{frac{l}{10}} implies l approx 1text{ m}$.
For a transverse wave on a string, the displacement is described by $y = Asin(kx – omega t)$. Then which of the following statement is incorrect?
The expression represents a wave propagating along the positive $x$-direction because of the term $(kx - omega t)$. The medium particles oscillate along the $y$-axis.
Velocity of sound in air is $320text{ m/s}$. If frequency of $1^{text{st}}$ overtone of a closed organ pipe is $480text{ Hz}$, then the length of the organ pipe is
For a closed organ pipe, the $1^{text{st}}$ overtone frequency is $f = frac{3v}{4L}$. Substituting the values: $480 = frac{3 times 320}{4L} implies L = 0.5text{ m} = 50text{ cm}$.
A stretched string of length $1text{ m}$ fixed at both ends having a mass of $10^{-4}text{ kg}$ is under a tension of $16text{ N}$. The speed of the transverse wave on the string would be
The velocity is given by $v = sqrt{frac{T}{mu}}$, where $mu = frac{M}{L} = 10^{-4}text{ kg/m}$. Thus, $v = sqrt{frac{16}{10^{-4}}} = 400text{ m/s}$.
A body cools from $80^circtext{C}$ to $70^circtext{C}$ in $12text{ minutes}$ and from $70^circtext{C}$ to $60^circtext{C}$ in time $t$ minutes. The value of $t$ is (temperature of surrounding is $40^circtext{C}$)
Using Newton's law of cooling: $frac{T_1 - T_2}{t} = Kleft(frac{T_1 + T_2}{2} - T_sright)$. From the first interval, we get $K = frac{1}{42}$. For the second interval, we obtain $t = 16.8text{ minutes}$.
A solid cube of side $4text{ m}$ having coefficient of areal expansion $2 times 10^{-5}/^circtext{C}$. If temperature is changed by $40^circtext{C}$ then the change in side length of the cube will be
The coefficient of linear expansion is $alpha = frac{beta}{2} = 10^{-5}/^circtext{C}$. Change in side length is $Delta L = L alpha Delta T = 4 times 10^{-5} times 40 = 1.6text{ mm}$.
Statement I: Internal energy of an ideal gas remains constant in an adiabatic process.
Statement II: In an adiabatic process, change in internal energy of a gas is equal to work done on or by the gas in the process.
In an adiabatic process, $Q = 0$ so $Delta U = -W$. Hence, internal energy changes (Statement I is incorrect) and the change equals work done (Statement II is correct).