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
[AIEEE-2009]
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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
[AIEEE-2011]
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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) $\mathrm{g} \mathrm{H}=(\mathrm{n}-2)^{2} \mathrm{u}^{2}$
[JEE-MAIN-2014]
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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 ?
[JEE Mains – 2017]
[JEE Mains – 2017]
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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.
[JEE Mains – 2018]
[JEE Mains – 2018]
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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)
fuck exams
these are not enough for practise but uploads some more questions
i really need of more privious questions
Yes
Very good but would have been better if updated
Well, even the website’s name is ‘e-Saral’. Way too easy.
more and harder questions please but good
Yes, indeed. These questions are way too easy! A better set of questions would do the trick.
Well/but not hard is pick up/plz update harder omes🙏🙏🙏🙏
Yes
How can। idownload it
Please update this
I want more previous year questions for jee mains
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I want more queries
Ya strength of questions are less but helpfull
Thanks esaral these questions are helpful for us
binod
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good questions
tthnk you but just 5 questions!!!!
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couldn’t understand last one alone
Nice but less questions
Yah you are right
Nice
Good but questions are less
Very helpful but not much helpful…
For practice
fine but very less questions
Xe
good questions could u provide PDF of it
fuckingly less questions
so less questions
Very less questions
Thank you for providing such a good questions
Please can you make a pdf of it.
great effort
Best
Yes
SUPER
hello I am preparing for jee pls help me in physics work power and energy
too few questions
Very helpful. Thank you!!
Plz give more questions.
Can’t u give more questions?
Good job
Plz give more questions..
sir we want to download direct pdf please provide
I want to download questions in PDF how to download
Good question
In This there are less Qestions
Nice