Calculus – Page 2

Particle in a tube

January 16th, 2022jeefirst0 Comments

JEE Advanced 2019 Paper 2, Question 3

A small particle of mass $m$ moving inside a heavy, hollow and straight tube along the tube axis undergoes elastic collision at two ends. The tube has no friction and it is closed at one end by a flat surface while the other end is fitted with a heavy movable flat piston as shown in figure. When the distance of the piston from the closed end is $L=L_{0}$ the particle speed is $v=v_{0} .$ The piston is moved inward at a very low speed $V$ such that $V \ll \frac{{\rm d} L}{L} v_{0}$ , where ${\rm d} L$ is the infinitesimal displacement of the piston. Which of …

Rod heated by wire

December 30th, 2021jeefirst0 Comments

JEE Advanced 2019 Paper 1, Question 3

A current carrying wire heats a metal rod. The wire provides a constant power $P$ to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature $T$ in the metal rod changes with time $t$ as

(1) $\begin{equation*} T(t) = T_0 (1 + \beta t^{\frac{1}{4}}) \end{equation*}$

where $\beta$ is a constant with appropriate dimension while $T_0$ is a constant with dimension of temperature. The heat capacity of the metal is,

$\frac{4 P [T(t) - T_0]^3}{\beta^4 T_0^4}$
$\frac{4 P [T(t) - T_0]^4}{\beta^4 T_0^5}$
$\frac{4 P [T(t) - T_0]^2}{\beta^4 T_0^3}$
$\frac{4 P [T(t) - T_0]}{\beta^4 T_0^2}$

Solution

The heat capacity of an object is defined by the relation

(2) $\begin{equation*} \Delta Q = C \Delta T \end{equation*}$

where $\Delta Q$ is the heat that the object absorbs and $\Delta T$ is the resulting temperature change of …

Spherical gas cloud

December 30th, 2021jeefirst0 Comments

JEE Advanced 2019 Paper 1, Question 1

Consider a spherical gas cloud of mass density $\rho(r)$ in free space where $r$ is the radial distance from its center. The gaseous cloud is made of particles of equal mass $m$ moving in circular orbits about the common center with the same kinetic energy $K$ . The force acting on the particles is their mutual gravitational force. If $\rho(r)$ is constant in time, the particle number density $n(r) = \rho(r)/m$ is [ $G$ is the universal gravitational constant]