*Simulator***Previous Years AIEEE/JEE Mains Questions**

Q. A particle has an initial velocity of $3 \hat{i}+4 \hat{j}$ and an acceleration of $0.4 \hat{\mathrm{i}}+0.3 \hat{\mathrm{j}}$. Its speed after 10s is :-
(1) 7 units
(2) 8.5 units
(3) 10 units
(4) $7 \sqrt{2}$ units

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**Sol.**(4) v = u + at u = 3i + 4j + (0.4 i + 0.3 j) × 10 = 3i + 4j + 4i + 3i u = 7i + 7j $|\overrightarrow{\mathrm{u}}|=\sqrt{7^{2}+7^{2}}$ $=7 \sqrt{2}$

Q. An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by $\frac{\mathrm{dv}}{\mathrm{dt}}=-2.5 \sqrt{\mathrm{v}}$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be :-
(1) 4 s (2) 8 s (3) 1 s (4) 2 s

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**Sol.**(4) $\overrightarrow{\mathrm{a}}=\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}$ $\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=-2.5 \sqrt{\mathrm{v}}$ $\int \mathrm{d} \mathrm{v} \times \mathrm{v}^{-1 / 2}=\int-2.5 \mathrm{dt}$ $\frac{\mathrm{v}^{\frac{1}{2}}}{\frac{1}{2}}=-2.5 \mathrm{t}$ $2 \sqrt{\mathrm{v}}=-2.5 \mathrm{t}+\mathrm{c}$ $2 \sqrt{6.25}=-2.5 \mathrm{t}+\mathrm{c}$ 2 × 0.25 = c 5 = c $2 \sqrt{\mathrm{v}}=-\mathrm{kt}+5$ 0 = –kt + 5 kt = 5 $\mathrm{t}=\frac{5 \mathrm{sec}}{2.5}=2 \mathrm{sec}$

Q. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is :
(1) $2 \mathrm{g} \mathrm{H}=\mathrm{nu}^{2}(\mathrm{n}-2)$
(2) $\mathrm{g} \mathrm{H}=(\mathrm{n}-2) \mathrm{u}^{2}$
(3) $2 \mathrm{g} \mathrm{H}=\mathrm{n}^{2} \mathrm{u}^{2}$
(4) $\mathrm{g} \mathrm{H}=(\mathrm{n}-2)^{2} \mathrm{u}^{2}$

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**Sol.**(1) Time to reach highest point $=\mathrm{t}=\frac{\mathrm{u}}{\mathrm{g}}$ time to reach ground = nt $\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$ $-\mathrm{H}=\mathrm{u}(\mathrm{nt})-\frac{1}{2} \mathrm{g}(\mathrm{nt})^{2}$ $\Rightarrow 2 \mathrm{gH}=\mathrm{nu}^{2}(\mathrm{n}-2)$

Q. A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time ?

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**Sol.**(1) Velocity at any time t is given by v = u + at v = v0 + (–g)t $\mathrm{v}=\mathrm{v}_{0}-\mathrm{gt}$ $\Rightarrow$ straight line with negative slope

Q. All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

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**Sol.**(1) In this question option (2) and (4) are the corresponding position – time graph and velocity –position graph of option (3) and its distance – time graph is given as Hence incorrect graph is option (1)

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Badiya

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Yes

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I want more queries

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Thanks esaral these questions are helpful for us

binod

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For practice

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Please can you make a pdf of it.

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too few questions

Very helpful. Thank you!!

Plz give more questions.

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Plz give more questions..

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I want to download questions in PDF how to download

Good question

In This there are less Qestions

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