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Two bodies of mass 1kg and 3kg have position vectors $\hat{i} + 2 \hat{j} + \hat{k}$ and $- 3 \hat{i} - 2 \hat{j} + \hat{k}$ , respectively. The centre of mass of this system has a position vector -

$- 2 \hat{i} + 2 \hat{k}$

$- 2 \hat{i} - \hat{j} + \hat{k}$

$2 \hat{i} - \hat{j} - 2 \hat{k}$

$- \hat{i} + \hat{j} + \hat{k}$

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Explanation

The position vector of centre of mass

$\vec{r} = \frac{m_{1} \vec{r_{1}} + m_{2} \vec{r_{2}}}{m_{1} + m_{2}}$

= $\frac{1 (\hat{i} + 2 \hat{j} + \hat{k}) + 3 (- 3 \hat{i} - 2 \hat{j} + \hat{k})}{1 + 3}$

= $\frac{1}{4} (- 8 \hat{i} - 4 \hat{j} + 4 \hat{k})$

$= - 2 \hat{i} - \hat{j} + \hat{k}$

The centre of mass changes its position only under the translatory motion. There is no effect of rotatory motion on centre of mass of the body.

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Four identical thin rods each of mass M and length t, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is

$\frac{4}{3} M t^{2}$

$\frac{2}{3} M t^{2}$

$\frac{13}{3} M t^{2}$

$\frac{1}{3} M t^{2}$

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Explanation

Apply theorem of parallel axisand the total moment of interia will be the sum of moment of inertia of each rod.

Moment of inertia of rod about an axis through its centre of mass and perpendicular to rod + (mass of rod) $\times$ (perpendicular distance between two axes)

= $\frac{M t^{2}}{12} + M {(\frac{t}{2})}^{2} = \frac{M t^{2}}{3}$

Moment of inertia of the system = $\frac{M t^{2}}{3} \times 4$

= $\frac{4}{3} M t^{2}$

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A shell of mass 200g is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. The initial velocity of the shell is -

100 $m s^{- 1}$

80 $m s^{- 1}$

40 $m s^{- 1}$

120 $m s^{- 1}$

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Explanation

In the given problem conservation of linear of momentum yields.

$m_{1} v_{1} + m_{2} v_{2} = 0$

or $4 v_{1} + 0.2 v_{2} = 0 . . . . (i)$

Conservation of energy yields.

$\frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} = 1050 0 r \frac{1}{2} \times 4 v_{1}^{2} + \frac{1}{2} \times 0.2 \times v_{2}^{2} = 1050 o r 2 v_{1}^{2} + 0.1 v_{2}^{2} = 1050 . . . . (i i)$

Solving Eqs. (i) and (ii) , we have

$v_{2} = 100 m / s$

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A thin rod of length L and mass M is bent at its midpoint into two halves so that the angle between them is $90^{°}$ . The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is

$\frac{M L^{2}}{24}$

$\frac{M L^{2}}{12}$

$\frac{M L^{2}}{6}$

$\frac{\sqrt{2} M L^{2}}{24}$

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Explanation

To find the moment of inertia of a bent rod, we need to consider the moment of inertia of each half of the rod about the bending point and then add them. The moment of inertia of a rod about a perpendicular axis through its midpoint is (ML^2)/12.

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A satellite is moving very close to a planet of density $ρ$ . The time period of the satellite is:

$\sqrt{\frac{3 π}{ρG}}$

${(\frac{3 π}{ρG})}^{3 / 2}$

$\sqrt{\frac{3 π}{2 ρG}}$

${(\frac{3 π}{2 ρG})}^{3 / 2}$

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Explanation

$T = \frac{2 πR}{v} = \frac{2 πR}{\sqrt{\frac{G M}{R}}} = \frac{2 {πR}^{\frac{3}{2}}}{\sqrt{G M}} T = \frac{2 π R^{\frac{3}{2}}}{\sqrt{G \frac{4}{3} {πR}^{3} ρ}} = \frac{2 {πR}^{\frac{3}{2}}}{2 \sqrt{\frac{π}{3} ρ G} R^{\frac{3}{2}}} = \sqrt{\frac{3 π}{ρ G}}$

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A projectile is fired upwards from the surface of the earth with a velocity ${kv}_{e}$ where $v_{e}$ is the escape velocity and k < 1. If r is the maximum distance from the center of the earth to which it rises and R is the radius of the earth, then r equals

$\frac{R}{k^{2}}$

$\frac{R}{1 - k^{2}}$

$\frac{2 R}{1 - k^{2}}$

$\frac{2 R}{1 + k^{2}}$

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Explanation

$(2) K_{i} + U_{i} = K_{f} + U_{f} \frac{1}{2} m {(k v_{e})}^{2} - \frac{G M m}{R} = 0 - \frac{G M m}{r} \frac{1}{2} m {(k \sqrt{\frac{2 G M}{R}})}^{2} - \frac{G M m}{R} = 0 - \frac{G M m}{r} \frac{k^{2}}{R} - \frac{1}{R} = - \frac{1}{r} r = \frac{R}{1 - k^{2}}$

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The gravitational potential difference between the surface of a planet and 10 m above is 5 J/kg. If the gravitational field is supposed to be uniform, the work done in moving a 2 kg mass from the surface of the planet to a height of 8 m is

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Explanation

$P o t e n t i a l d i f f e r e n c e a t 10 m = 5 J / k g P o t e n t i a l d i f f e r e n c e a t 8 m = \frac{5}{10} \times 8 = 4 J / k g W o r k d o n e = m ∆ V = 2 \times 4 = 8 J$

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A planet is moving in an elliptical orbit. If T, V, E, and L stand, respectively, for its kinetic energy, gravitational potential energy, total energy and angular momentum about the center of orbit, then

T is conserved

V is always positive

E is always negative

the magnitude of L is conserved but its direction changes continuously

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Explanation

The direction of angular momentum does not change. It remains same at each and every point of the path. We can find the direction of angular momentum using right hand thumb rule.

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In planetary motion the areal velocity of the position vector of a planet depends on the angular velocity ( $ω$ ) and the distance of the planet from the sun (r). The correct relation for areal velocity is:

$\frac{d A}{d t} α ω r$

$\frac{d A}{d t} α ω^{2} r$

$\frac{d A}{d t} α ω r^{2}$

$\frac{d A}{d t} α \sqrt{ω r}$

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Explanation

$(3) A = {πr}^{2} \frac{dA}{dt} = 2 πr \frac{dr}{dt} = 2 πrv \frac{dA}{dt} = 2 {πr}^{2} ω$

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If A is the areal velocity of a planet of mass M, its angular momentum is

$\frac{M}{A}$

2MA

$A^{2} M$

${AM}^{2}$

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Explanation

Angular momentum

= 2 x mass x Areal speed = 2 MA

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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.