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Language English हिंदी

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Three masses are placed on the x-axis:

300 g at origin, 500 g at x= 40 cm and 400g

at x=70 cm. The distance of the center of

mass from the origin is

40 cm

45 cm

50 cm

30 cm

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Explanation

$x_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}} = \frac{300 (0) + 500 (40) + 400 (70)}{300 + 500 + 400} = \frac{20000 + 28000}{1200} = 40 cm$

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The instantaneous angular position of a point on

a rotating wheel is given by the equation

$θ (t) = 2 t^{3} - 6 t^{2}$

The torque on the wheel becomes zero at

t=0.5 s

t=0.25 s

t=2 s

t=1 s

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Explanation

According to question, torque, $τ = 0$

It means that, $α = 0$

$α = \frac{d^{2} θ}{{dt}^{2}}$

Given $θ (t) = 2 t^{3} - 6 t^{2}$

so, $\frac{dθ}{dt} = 6 t^{2} - 12 t \frac{d^{2} θ}{{dt}^{2}} = 12 t - 12 (since, α = \frac{d^{2} θ}{{dt}^{2}} = 0) 12 t - 12 = 0 t = 1 s$

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The moment of inertia of a thin uniform rod of

mass M and length L about an axis passing

through its mid-point and perpendicular to its

length is $I_{0}$ . Its moment of inertia about an

axis passing through one of its ends and

perpendicular to its length is

$I_{0} + {ML}^{2} / 4$

$I_{0} + 2 {ML}^{2}$

$I_{0} + {ML}^{2}$

$I_{0} + {ML}^{2} / 2$

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Explanation

The theorem of parallel axis for moment of

inertia.

$I = I_{CM} + {Mh}^{2} I = I_{0} + M {(\frac{L}{2})}^{2} I = I_{0} + \frac{{ML}^{2}}{4}$

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A body of mass M hits normally a rigid wall with velocity v and bounces back with the same velocity. The impulse experienced by the body is

1.5 Mv

2 Mv

zero

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Explanation

Impulse $| J | = | ∆ p |$

$= M v - (- M v) = 2 M v$

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A mass m moving horizontally (along the x-axis) with velocity v collides and sticks to mass of 3m moving vertically upward (along the y-axis) with velocity 2v. The final velocity of the combination is

$\frac{1}{4} v \hat{i} + \frac{3}{2} v \hat{j}$

$\frac{1}{3} v \hat{i} + \frac{2}{3} v \hat{j}$

$\frac{2}{3} v \hat{i} + \frac{1}{3} v \hat{j}$

$\frac{3}{2} v \hat{i} + \frac{1}{4} v \hat{j}$

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A circular disk of moment of inertia $I_{t}$ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed $ω_{i} .$ Another disk of moment of inertia $I_{b}$ is dropped coaxially onto the rotation disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed $ω_{f} .$ The energy lost by the initially rotating disc to friction is

$\frac{1}{2} \frac{I_{b}^{2}}{(I_{t} + I_{b})} ω_{i}^{2}$

$\frac{1}{2} \frac{I_{t}^{2}}{(I_{t} + I_{b})} ω_{i}^{2}$

$\frac{1}{2} \frac{I_{b} - I_{t}}{(I_{t} + I_{b})} ω_{i}^{2}$

$\frac{1}{2} \frac{I_{b} I_{t}}{(I_{t} + I_{b})} ω_{i}^{2}$

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Explanation

Loss of energy, $∆ E = \frac{1}{2} I_{t} ω_{i}^{2} - \frac{1}{2} \frac{I_{t}^{2} ω_{i}^{2}}{(I_{t} + I_{b})}$

= $\frac{1}{2} \frac{I_{b} I_{t} ω_{i}^{2}}{(I_{t} + I_{b})}$

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A man of 50 kg mass is standing in a gravity free space at a height of 10m above the floor. He throws a stone of 0.5 kg mass downwards with a speed $2 m s^{- 1} .$ When the stone reaches the floor, the distance of the man above the floor will be

9.9m

10.1m

10m

20m

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Explanation

mr = constant

$m_{1} r_{1} = m_{2} r_{2}$

$\Rightarrow r_{2} = \frac{m_{1} r_{1}}{m_{2}}$

= $\frac{0.5 \times 10}{50} = 0.1$

The distance of the man above the floor (total height)=10+0.1=10.1m.

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An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, 1 kg first part moving with a velocity of $12 {ms}^{- 1}$ and 2 kg second part moving with a velocity of $8 {ms}^{- 1} .$ If the third part flies off with a velocity of $4 {ms}^{- 1},$ its mass would be

$5 kg$

$7 kg$

$17 kg$

$3 kg$

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If $\vec{F}$ is the force acting on a particle having position vector $\vec{r}$ and $\vec{τ}$ be the torque of this force about the origin, then

$\vec{r} . \vec{τ} \neq 0 and \vec{F} . \vec{τ} = 0$

$\vec{r} . \vec{τ} > 0 and \vec{F} . \vec{τ} < 0$

$\vec{r} . \vec{τ} = 0 and \vec{F} . \vec{τ} = 0$

$\vec{r} . \vec{τ} = 0 and \vec{F} . \vec{τ} \neq 0$

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Explanation

Key Idea Torque is an axial vector ie, its direction is always perpendicular to the plane containing vectors $\vec{r}$ and $\vec{F}$ .

$\vec{τ} = \vec{r} \times \vec{F}$

Torque is perpendicular to both $\vec{r}$ and $\vec{F}$

∴ $\vec{τ} \cdot \vec{r} = 0$ $\vec{F} \cdot \vec{τ} = 0$

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A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity $ω .$ If two objects each of mass m be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity -

$\frac{ω (M - 2 m)}{M + 2 m}$

$\frac{ω M}{M + 2 m}$

$\frac{ω (M + 2 m)}{M}$

$\frac{ω M}{M + m}$

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Explanation

Key Idea

Apply law of conservation of angular momentum.

$I_{1} ω_{1} = I_{2} ω_{2}$

In the given case

$I_{1} = M R^{2}$

$I_{2} = M R^{2} + 2 m R^{2}$

$ω_{1} = ω$

Then $ω_{2} = \frac{I_{1}}{I_{2}} ω = \frac{M}{M + 2 m} ω$

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Yes. You can attempt every NEET question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.