December 2021 – JEE First

Flux from a charged shell

December 30th, 2021jeefirst0 Comments

JEE Advanced 2019 Paper 1, Question 8

A charged shell of radius $R$ carries a total charge $Q$ . Given $\Phi$ as the flux of electric field through a closed cylindrical surface of height $h$ , radius $r$
and with its center same as that of the shell. Here, the center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following option(s) is/are correct?

[ $\epsilon_0$ is the permittivity of free space]

If $h > 2R$ and $r > R$ then $\Phi = Q/\epsilon_0$
If $h < 8R/5$ and $r = 3R/5$ then $\Phi = 0$
If $h > 2R$ and $r = 3R/5$ then $\Phi = Q/5\epsilon_0$
If $h > 2R$ and $r = 4R/5$ then $\Phi = Q/5\epsilon_0$

Related problems:
Electric field from a sphere …

Conducting wire in a magnetic field

December 30th, 2021jeefirst0 Comments

JEE Advanced 2019 Paper 1, Question 6

A conducting wire of parabolic shape, initially $y = x^2$ , is moving with velocity $\vec v = v_0 \hat i$ in a non-uniform magnetic field $\vec B = B_0 \left( 1 + \left( \frac{y}{L} \right)^\beta \right) \hat k$ , as shown in the figure below. If $v_0, B_0, L$ and $\beta$ are positive constants and $\Delta \phi$ is the potential difference developed between the ends of the wire, then the correct statement(s) is/are:

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$|\Delta \phi| = \frac{1}{2} B_0 v_0 L$ for $\beta = 0$ .
$|\Delta \phi| = \frac{4}{3} B_0 v_0 L$ for $\beta = 2$ .
$|\Delta \phi|$ remains the same if the parabolic wire is replaced by a straight wire, $y=x$ initially, of length $\sqrt{2} L$ .
$|\Delta \phi|$ is proportional to the length of the wire projected on the $y$ axis.

Radioactive decay of Potassium

December 30th, 2021jeefirst0 Comments

JEE Advanced 2019 Paper 1, Question 4

In a radioactive sample, ${}^{40}_{19}{\rm K}$ nuclei decay into stable ${}^{40}_{20}{\rm Ca}$ nuclei with decay constant $4.5 \times 10^{-10}$ per year or into stable ${}^{40}_{18}{\rm Ar}$ nuclei with decay constant $0.5 \times 10^{-10}$ per year. In this sample all the stable ${}^{40}_{20}{\rm Ca}$ and ${}^{40}_{18}{\rm Ar}$ nuclei are produced by the ${}^{40}_{19}{\rm K}$ nuclei only. In time $t_1 \times 10^9$ years, if the ratio of the sum of stable ${}^{40}_{20}{\rm Ca}$ and ${}^{40}_{18}{\rm Ar}$ nuclei to the radioactive ${}^{40}_{19}{\rm K}$ nuclei is 99, the value of $t_1$ will be [Given $\ln 10 = 2.3$ ],

1.15
9.2
2.3
4.6

Solution

Let us first consider the situation where a sample of $X$ nuclei decays into a single type of nuclei $Y$ . The number $N$ of $X$ nuclei changes with …

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 …

Insulating spherical shell with a hole

December 30th, 2021jeefirst0 Comments

JEE Advanced 2019 Paper 1, Question 2.

A thin spherical insulating shell of radius $R$ carries a uniformly distributed charge such that the potential at its surface is $V_0$ . A hole with a small area $\alpha 4 \pi R^2 \ (\alpha \ll 1)$ is made on the shell without affecting the rest of the shell. Which one of the following statements is correct?

The potential at the center of the shell is reduced by $2 \alpha V_0$
The magnitude of the eletric field at the center of the shell is reduced by $\frac{\alpha V_0}{2 R}$
The ratio of the potential at the center of the shell to that of the point at $R/2$ from center towards

…