SECTION - B [PHYSICS - NUMERIC] - JEE STD 11 Science Questions

Question 14 Marks

The maximum speed of a boat in still water is $27 \mathrm{~km} / \mathrm{h}$. Now this boat is moving downstream in a river flowing at $9 \mathrm{~km} / \mathrm{h}$. A man in the boat throws a ball vertically upwards with speed of $10 \mathrm{~m} / \mathrm{s}$. Range of the ball as observed by an observer at rest on the river bank, is ____________ cm . (Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ )

Answer

(2000)
Sol.

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Question 24 Marks

In a hydraulic lift, the surface area of the input piston is $6 \mathrm{~cm}^{2}$ and that of the output piston is $1500 \mathrm{~cm}^{2}$. If 100 N force is applied to the input piston to raise the output piston by 20 cm , then the work done is____________ kJ.

Answer

(5)
Sol.

$\frac{\mathrm{F}_{1}}{\mathrm{~A}_{1}}=\frac{\mathrm{F}_{2}}{\mathrm{~A}_{2}}, \frac{100}{6}=\frac{\mathrm{F}}{1500}, \mathrm{~F}=\frac{50}{3} \times 1500$
$\mathrm{F}=50 \times 500=25 \times 10^{3} \mathrm{~N}$
$\omega=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{S}}=25 \times 10^{3} \times \frac{20}{100}$
$=5 \times 10^{3}=5 \mathrm{~kJ}$

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Question 34 Marks

Two light beams fall on a transparent material block at point 1 and 2 with angle $\theta_{1}$ and $\theta_{2}$, respectively, as shown in figure. After refraction, the beams intersect at point 3 which is exactly on the interface at other end of the block. Given : the distance between 1 and $2, d=4 \sqrt{3} \mathrm{~cm}$ and $\theta_{1}=\theta_{2}=\cos ^{-1}\left(\frac{n_{2}}{2 n_{1}}\right)$, where refractive index of the block $\mathrm{n}_{2}>$ refractive index of the outside medium $n_{1}$, then the thickness of the block is
____________ cm .

Answer

(6)
Sol.

$\mathrm{n}_{1} \sin \left(90-\theta_{1}\right)=\mathrm{n}_{2} \sin \theta_{3}$
$\mathrm{n}_{1} \cos \theta_{1}=\mathrm{n}_{2} \sin \theta_{3}$
$\mathrm{n}_{1} \frac{\mathrm{n}_{2}}{2 \mathrm{n}_{1}}=\mathrm{n}_{2} \sin \theta_{3}$
$\frac{1}{2}=\sin \theta_{3}, \theta_{3}=30$
$\tan 30=\frac{\mathrm{d}}{2(\mathrm{t})}$
$\mathrm{t}=\frac{\mathrm{d} \sqrt{3}}{2}=\frac{4 \sqrt{3} \times \sqrt{3}}{2} \mathrm{~cm}=6 \mathrm{~cm}$

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Question 44 Marks

A container of fixed volume contains a gas at $27^{\circ} \mathrm{C}$. To double the pressure of the gas, the temperature of gas should be raised to ____________ ${ }^{\circ} \mathrm{C}$.

Answer

(327)
Sol.
$\frac{\mathrm{P}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2}}{\mathrm{~T}_{2}}$
$\frac{\mathrm{P}}{300}=\frac{2 \mathrm{P}}{\mathrm{T}_{2}}$
$\mathrm{T}_{2}=600 \mathrm{~K}$
$\mathrm{T}_{2}=327^{\circ} \mathrm{C}$

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Question 54 Marks

The coordinates of a particle with respect to origin in a given reference frame is $(1,1,1)$ meters. If a force of $\vec{F}=\hat{i}-\hat{j}+\hat{k}$ acts on the particle, then the magnitude of torque (with respect to origin) in $z$-direction is ____________ .

Answer

(2)
Sol.
$\vec{\tau}=\overrightarrow{\mathbf{r}} \times \overrightarrow{\mathrm{F}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1\end{array}\right|$
$\vec{\tau}=\hat{k}(-1-1)=-2 \hat{k}$
$|\vec{\tau}|=2 \mathrm{Nm}$

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JEE SECTION - B [PHYSICS - NUMERIC]

SECTION - B [PHYSICS - NUMERIC]

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