If $\vec{F}$ is the force acting on a particle having position vector $\vec{r}$ and $\vec{\tau}$ be the torque of this force about the origin, then: (2009)
Torque is defined as the cross product of position vector and force: $\vec{\tau} = \vec{r} \times \vec{F}$.
By the properties of cross products, the resulting vector $\vec{\tau}$ is perpendicular to both $\vec{r}$ and $\vec{F}$.
Therefore, their dot products must be zero: $\vec{r}\cdot\vec{\tau} = 0$ and $\vec{F}\cdot\vec{\tau} = 0$.
A thin circular ring of mass $M$ and radius $R$ is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity $\omega$. If two objects each of mass $m$ be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity: (2009)
By conservation of angular momentum, initial angular momentum equals final angular momentum ($I_1\omega_1 = I_2\omega_2$).
Initially, $I_1 = MR^2$ and $\omega_1 = \omega$. After attaching masses, $I_2 = MR^2 + 2mR^2 = (M+2m)R^2$.
Equating them: $MR^2\omega = (M+2m)R^2\omega' \implies \omega' = \frac{M\omega}{M + 2m}$.
A round disc of moment of inertia $I_2$ about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia $I_1$ rotating with an angular velocity $\omega$ about the same axis. The final angular velocity of the combination of discs is: (2004)
Since no external torque acts on the system, angular momentum is conserved.
Initial angular momentum $L_i = I_1\omega$.
Final angular momentum $L_f = (I_1 + I_2)\omega'$. Equating $L_i$ and $L_f$, we get $\omega' = \frac{I_1\omega}{I_1 + I_2} $.
What will be the formula of mass of the earth in terms of $g$, $R$ and $G$? (1996)
The acceleration due to gravity on the surface of the earth is given by $g = \frac{GM}{R^2}$. Rearranging this formula to solve for the mass of the earth $M$, we get $M = \frac{gR^2}{G}$.
Which of the following rods, (given radius $r$ and length $l$) each made of the same material and whose ends are maintained at the same temperature will conduct most heat? (2005)
The rate of heat conduction is $H = \frac{KA\Delta T}{l} = \frac{K(\pi r^2)\Delta T}{l}$. Since material and $\Delta T$ are same, $H \propto \frac{r^2}{l}$. This ratio is maximum for $r = 2r_0$ and $l = l_0$ (ratio $\propto 4$).
The quantities of heat required to raise the temperature of two solid copper spheres of radii $r_1$ and $r_2$ ($r_1 = 1.5 r_2$) through $1\text{ K}$ are in the ratio: (2020)
Heat required $Q = mc\Delta T = (\frac{4}{3}\pi r^3 \rho)c\Delta T$. For the same material and $\Delta T$, $Q \propto r^3$. Ratio $= (\frac{r_1}{r_2})^3 = (1.5)^3 = (\frac{3}{2})^3 = \frac{27}{8}$.
A piece of ice falls from a height $h$ so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of $h$ is [Latent heat of ice is $3.4 \times 10^5\text{ J/kg}$ and $g = 10\text{ N/kg}$]: (2016 – I)
Energy absorbed by ice $= \frac{1}{4} mgh$. This energy melts the ice, so $\frac{1}{4} mgh = mL$. Substituting values: $h = \frac{4L}{g} = \frac{4 \times 3.4 \times 10^5}{10} = 13.6 \times 10^4\text{ m} = 136\text{ km}$.
On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are $39^\circ\text{W}$ and $239^\circ\text{W}$ respectively. What will be the temperature on the new scale, corresponding to a temperature of $39^\circ\text{C}$ on the Celsius scale? (2008)
Using temperature scale relation: $\frac{W - 39}{239 - 39} = \frac{C - 0}{100 - 0}$. Substituting $C = 39$: $\frac{W - 39}{200} = \frac{39}{100} \Rightarrow W - 39 = 78 \Rightarrow W = 117^\circ\text{W}$.
Mercury thermometer can be used to measure temperature upto: (1992)
The boiling point of mercury is approximately $356.7^\circ\text{C}$. Therefore, a standard mercury thermometer can measure temperatures up to around $360^\circ\text{C}$.