Q141JEE Main 2003MCQ

A spring of spring constant $5 \times 10^3 \frac{\text{N}}{\text{m}}$ is stretched initially by $5 {\text{cm}}$ from the unstretched position. Then the work required to stretch it further by another $5 {\text{cm}}$ is

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📖 Explanation

Given $k=5 \times 10^3 \frac{\text{N}}{\text{m}}$ Work done when a spring stretched from ${\text{x}}_1 {\text{cm}}$ to ${\text{x}}_2 {\text{cm}}$ , $\;W=\;\frac{1}{2} k (x_2^{2-}x_1^2)$ $\;=\;\frac{1}{2} \times 5 \times 10^3 [(0.1)^{2-}(0.05)^2]$ $\;=\;\frac{5000}{2} \times 0.15 \times 0.05=18.75 {\text{Nm}}$

Q142JEE Main 2003MCQ

A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time ${ }^{'} t$ ' is proportional to

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📖 Explanation

We know that $F \times v=$ Power According to the question, power is constant. $\therefore F \times v=c \;\;$ where $c=$ constant $\therefore m \;\frac{d v}{d t} \times v=c \;\; ( \therefore F=m a=\;\frac{m d v}{d t})$ $\therefore m ∫_{0}^{v} v d v=c ∫_{0}^{t} \text{dt} \;\; \therefore \;\frac{1}{2} m v^2=c t$ $\therefore v=\sqrt{\;\frac{2 c}{m}} \times t^{1 / 2}$ $\;\frac{d x}{d t}=\sqrt{\;\frac{2 c}{m}} \times t^{1 / 2} \;\;$ where $v=\;\frac{d x}{d t}$ $\therefore ∫_{0}^{x} \text{dx}=\sqrt{\;\frac{2 c}{m}} \times ∫_{0}^{t} t^{1 / 2} \text{dt}$ $x=\sqrt{\;\frac{2 c}{m}} \times \;\frac{2 t^{5 / 2}}{3} \Rightarrow x \propto t^{3 / 2}$

Q143JEE Main 2003MCQ

A wire suspended vertically from one of its ends is stretched by attaching a weight of $200 {\text{N}}$ to the lower end. The weight stretches the wire by $1 {\text{mm}}$ . Then the elastic energy stored in the wire is

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📖 Explanation

The elastic potential energy $\;=\;\frac{1}{2} \times \;\text{ Force }\; \times \;\text{ extension }\;$ $\;=\;\frac{1}{2} \times 200 \times 0.001=0.1 {\text{J}}$

Q144JEE Main 2002MCQ

A ball whose kinetic energy E, is projected at an angle of $45^{\circ}$ to the horizontal. The kinetic energy of the ball at the highest point of its height will be

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📖 Explanation

Assume the ball of mass $m$ is projected with a speed $u$ . Then the kinetic energy(E) at the point of projection $=\;\frac{1}{2} m u^2$ At highest point of flight only horizontal component of velocity $u \cos \theta$ present as at highest point vertical component of velocity is $=0$ . Note : The horizontal component of velocity does not change in entire projectile motion. At highest point the velocity is $=u \cos \theta =u \cos 45^{\circ}=\;\frac{u}{\sqrt{2}}$ $\therefore$ The kinetic energy at the height point $=\;\frac{1}{2} m(\;\frac{u}{\sqrt{2}})^2$ $=\;\frac{1}{2} m u^2 \times \;\frac{1}{2}=\;\frac{E}{2}$

Q145JEE Main 2002MCQ

If a body looses half of its velocity on penetrating $3 {\text{cm}}$ in a wooden block, then how much will it penetrate more before coming to rest?

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📖 Explanation

We know the work energy theorem, $W=\Delta K=F S$ For first penetration, by applying work energy theorem we get, $\;\frac{1}{2} m v^{2-}\;\frac{1}{2} m(\;\frac{v}{2})^2=F \times 3 . . .(i)$ For second penetration, by applying work energy theorem we get, $\;\frac{1}{2} m(\;\frac{v}{2})^{2-}0=F \times S$ ....(ii) On dividing $(i i)$ by $(i)$ $\;\frac{1 / 4}{3 / 4}=S / 3$ $\therefore S=1 {\text{cm}}$

Q146JEE Main 2002MCQ

A spring of force constant $800 \frac{\text{N}}{\text{m}}$ has an extension of $5 {\text{cm}}$ . The work done in extending it from 5 ${\text{cm}}$ to $15 {\text{cm}}$ is

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📖 Explanation

When we extend the spring by $d x$ then the work done $d W=k x d x$ Applying integration both sides we get, $\therefore W=k ∫_{0.05}^{0.15} x d x$ $=\;\frac{800}{2}[(0.15)^{2-}(0.05)^2]$ $=8 J$

Q147MCQ

A 60 HP electric motor lifts an elevator havinga maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N,the speed of the elevator at full load is close to:(1 HP = 746 W, g = 10 ms $^{-2}$ ) [7-Jan-2020 Shift 1]

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📖 Explanation

Let elevator is moving upward with constant speed V. Tension in cable $T = 2000 g + f_r$ $= 2000 + 4000$ T = 24000 NPower P = TV \Rightarrow 60 × 746 = (24000) V $\text{V}={60\times 746}/24000=1.865$ ≈ 1.9 m/s.

Q148NAT

If the maximum load carried by an elevator is $1400 {\text{kg}}$ ( $600 {\text{kg}}-$ Passengers $+800 {\text{kg}}$ - elevator), which is moving up with a uniform speed of $3 {\text{m}} {\ s}^{-1}$ and the frictional force acting on it is $2000 {\text{N}}$ , then the maximum power used by the motor is _______ ${\text{kW}} ( {\text{g}}=10 {\text{m}} / {\ s}^2)$ [10-Apr-2023 shift 2]

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📖 Explanation

Tension in the string $\Rightarrow 16000 {\text{N}}$ Maximum power $=( {\text{F}})( {\text{V}})$ $\;=16000 \times 3$ $\;=48000$ $\;=48 {\text{kW}}$ Ans. 48