1
JEE Main 2022 (Online) 29th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A ball is released from a height h. If $$t_{1}$$ and $$t_{2}$$ be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between $$t_{1}$$ and $$t_{2}$$.

A
$$t_{1}=(\sqrt{2}) t_{2}$$
B
$$t_{1}=(\sqrt{2}-1) t_{2}$$
C
$$t_{2}=(\sqrt{2}+1) t_{1}$$
D
$$t_{2}=(\sqrt{2}-1) t_{1}$$
2
JEE Main 2022 (Online) 29th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height $$\frac{h}{3}$$ while going up and coming down respectively.

A
$$\frac{\sqrt{2}-1}{\sqrt{2}+1}$$
B
$$\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$$
C
$$\frac{\sqrt{3}-1}{\sqrt{3}+1}$$
D
$$\frac{1}{3}$$
3
JEE Main 2022 (Online) 29th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If $$\mathrm{t}=\sqrt{x}+4$$, then $$\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)_{\mathrm{t}=4}$$ is :

A
4
B
zero
C
8
D
16
4
JEE Main 2022 (Online) 28th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

At time $$t=0$$ a particle starts travelling from a height $$7 \hat{z} \mathrm{~cm}$$ in a plane keeping z coordinate constant. At any instant of time it's position along the $$\hat{x}$$ and $$\hat{y}$$ directions are defined as $$3 \mathrm{t}$$ and $$5 \mathrm{t}^{3}$$ respectively. At t = 1s acceleration of the particle will be

A
$$-30 \hat{y}$$
B
$$30 \hat{y}$$
C
$$3 \hat{x}+15 \hat{y}$$
D
$$3 \hat{x}+15 \hat{y}+7 \hat{z}$$
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