1
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+1
-0

The graph of stopping potential ' $\mathrm{V}_{\mathrm{s}}$ ' against frequency ' $v$ ' of incident radiation is plotted for two different metals ' X ' and ' Y ' as shown in graph. ' $\phi_x$ ' and ' $\phi_y$ ' are work functions of ' $x$ ' and ' $Y$ ' respectively then

MHT CET 2024 3rd May Evening Shift Physics - Dual Nature of Radiation Question 53 English

A
$\phi_x=\phi_y$
B
$\phi_x<\phi_y$
C
$\phi_x>\phi_y$
D
$\phi_x=\phi_y=0$
2
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+1
-0

A steel ball of radius 6 mm has a terminal speed of $12 \mathrm{cms}^{-1}$ in a viscous liquid. What will be the terminal speed of a steel ball of radius 3 mm in the same liquid?

A
$12 \mathrm{~cms}^{-1}$
B
$9 \mathrm{~cms}^{-1}$
C
$6 \mathrm{~cms}^{-1}$
D
$3 \mathrm{~cms}^{-1}$
3
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+1
-0

Two coils are kept near each other. When no current passess through first coil and current in the $2^{\text {nd }}$ coil increases at the rate $10 \mathrm{~A} / \mathrm{s}$, the e.m.f. in the $1^{\mathbb{P}}$ coil is 20 mV . When no current passes through $2^{\text {nd }}$ coil and 3.6 A current passes through $1^2$ coil the flux linkage in coil 2 is

A
$1.2 \times 10^{-3} \mathrm{~Wb}$
B
$1.8 \times 10^{-3} \mathrm{~Wb}$
C
$3.6 \times 10^{-3} \mathrm{~Wb}$
D
$7.2 \times 10^{-3} \mathrm{~Wb}$
4
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+1
-0

Two rods, one of copper ( Cu$)$ and the other of iron ( Fe ) having initial lengths $\mathrm{L}_1$ and $\mathrm{L}_2$ respectively are connected together to form a single rod of length $L_1+L_2$. The coefficient of linear expansion of Cu and Fe are $\alpha_c$ and $\alpha_i$ respectively. If the length of each rod increases by the same amount when their temperatures are raised by $t^{\circ} \mathrm{C}$, then ratio of $\frac{L_1-L_2}{L_1+L_2}$ will be

A
$\frac{\alpha_i}{\alpha_c+\alpha_i}$
B
$\frac{\alpha_c}{\alpha_c+\alpha_i}$
C
$\frac{\alpha_i-\alpha_c}{\alpha_c+\alpha_i}$
D
$\frac{\alpha_c-\alpha_i}{\alpha_c+\alpha_i}$
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