A block ' A ' takes 2,s to slide down a frictionless incline of 30^{circ} and length ' l

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4-1.Newton's Laws of Motion

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A block '$A$' takes $2\,s$ to slide down a frictionless incline of $30^{\circ}$ and length ' $l$ ', kept inside a lift going up with uniform velocity ' $v$ '. If the incline is changed to $45^{\circ}$, the time taken by the block, to slide down the incline, will be approximately $........\,s$

A$2.66$

B$0.83$

C$1.68$

D$0.70$

(JEE MAIN-2022)

Solution

$a = g \sin \theta$
$\ell=\frac{1}{2} g \sin 30^{\circ}(2)^{2}$
$\ell=\frac{1}{2} g \sin 45^{\circ} t ^{2}$
$\left(\frac{1}{2}\right)(4)=\frac{1}{\sqrt{2}} t ^{2} \Rightarrow t =\sqrt{2 \sqrt{2}}=1.68$

Standard 11

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(AIEEE-2010)

A block '$A$' takes $2\,s$ to slide down a frictionless incline of $30^{\circ}$ and length ' $l$ ', kept inside a lift going up with uniform velocity ' $v$ '. If the incline is changed to $45^{\circ}$, the time taken by the block, to slide down the incline, will be approximately $........\,s$

Solution

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An object of mass $'m'$ is being moved with a constant velocity under the action of an applied force of $2\, {N}$ along a frictionless surface with following surface profile.
The correct applied force vs distance graph will be: