1
KCET 2025
MCQ (Single Correct Answer)
+1
-0

In an experiment to determine the figure of merit of a galvanometer by half deflection method, a student constructed the following circuit.

KCET 2025 Physics - Current Electricity Question 4 English

He unplugged a resistance of $5200 \Omega$ in R . When $\mathrm{K}_1$ is closed and $\mathrm{K}_2$ is open, the deflection observed in the galvanometer is 26 div. When $K_2$ is also closed and a resistance of $90 \Omega$ is removed in S , the deflection between 13 div. The resistance of galvanometer is nearly

A
$45.0 \Omega$
B
$103.0 \Omega$
C
$91.6 \Omega$
D
$116.0 \Omega$
2
KCET 2025
MCQ (Single Correct Answer)
+1
-0

While determining the coefficient of viscosity of the given liquid, a spherical steel ball sinks by a distance $\mathrm{h}=0.9 \mathrm{~m}$. The radius of the ball $\mathrm{r}=\sqrt{3} \times 10^{-3} \mathrm{~m}$. The time taken by the ball to sink in three trails are tabulated as follows.

$$ \begin{array}{|l|l|} \hline \text { Trial No. } & \text { Time taken by the ball to fall by } \mathrm{h} \text { (in second) } \\ \hline 1 . & 2.75 \\ \hline 2 . & 2.65 \\ \hline 3 . & 2.70 \\ \hline \end{array} $$

The difference between the densities of the steel ball and the liquid is $7000 \mathrm{~kg} \mathrm{~m}^{-3}$. If $\mathrm{g}=10 \mathrm{~ms}^{-2}$, then the coefficient of viscosity of the given liquid at room temperature is

A
0.14 Pa.s
B
$0.14 \times 10^{-3} \mathrm{~Pa} . \mathrm{s}$
C
14 Pa.s
D
0.28 Pa.s
3
KCET 2025
MCQ (Single Correct Answer)
+1
-0
Which of the following expression can be deduced on the basis of dimensional analysis? (All symbols have their usual meanings)
A
$\mathrm{x}=\mathrm{A} \cos \omega \mathrm{t}$
B
$\mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}}$
C
$\mathrm{F}=6 \pi \eta \mathrm{r} \nu$
D
$\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}{ }^2$
4
KCET 2025
MCQ (Single Correct Answer)
+1
-0
Two stones begin to fall from rest from the same height, with the second stone starting to fall ' $\Delta \mathrm{t}$ ' seconds after the first falls from rest. The distance of separation between the two stones becomes ' H ', ' $\mathrm{t}_0$ ' seconds after the first stone starts its motion. Then $\mathrm{t}_0$ is equal to
A
$\frac{\mathrm{H}}{\Delta \mathrm{t}}+\frac{\Delta \mathrm{t}}{2 \mathrm{~g}}$
B
$\frac{\mathrm{H}}{\mathrm{g} \Delta \mathrm{t}}-\frac{\Delta \mathrm{t}}{2}$
C
$\frac{\mathrm{H}}{\mathrm{g} \Delta \mathrm{t}}+\frac{\Delta \mathrm{t}}{2}$
D
$\frac{\mathrm{H}}{\mathrm{g} \Delta \mathrm{t}}$
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