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NEET NEET Questions — Free MCQ Practice

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A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to –K/r², where K is a constant. The total energy of the particle is

A.

$\frac{K}{2 r}$

B.

$- \frac{K}{2 r}$

C.

$- \frac{K}{r}$

D.

$\frac{K}{r}$

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Explanation

Here $\frac{m v^{2}}{r} = \frac{K}{r^{2}}$

∴ K.E. $= \frac{1}{2} m v^{2} = \frac{K}{2 r}$

$U = - \int_{\infty}^{r} F . d r = - \int_{\infty}^{r} (- \frac{K}{r^{2}}) d r = - \frac{K}{r}$

Total energy $E = K.E. + P.E. = \frac{K}{2 r} - \frac{K}{r} = - \frac{K}{2 r}$

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The displacement x of a particle moving in one dimension under the action of a constant force is related to the time t by the equation $t = \sqrt{x} + 3$ , where x is in meters and t is in seconds. The work done by the force in the first 6 seconds is

A.

9 J

B.

6 J

C.

0 J

D.

3 J

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Explanation

$x = {(t - 3)}^{2}$ ⇒ $v = \frac{d x}{d t} = 2 (t - 3)$

at $t = 0$ ; $v_{1} = - 6 m / s$ and at $t = 6 \sec$ , $v_{2} = 6 m / s$

so, change in kinetic energy $= W = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2} = 0$

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A force $F = - K (y i + x j)$ (where K is a positive constant) acts on a particle moving in the xy-plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force F on the particles is

A.

$- 2 K a^{2}$

B.

$2 K a^{2}$

C.

$- K a^{2}$

D.

$K a^{2}$

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Explanation

While moving from (0,0) to (a,0)

Along positive x-axis, y = 0

$∴ \vec{F} = - k x \hat{j}$

i.e. force is in negative y-direction while displacement is in positive x-direction.

∴ W₁ = 0

Because force is perpendicular to displacement

Then particle moves from (a, 0) to (a, a) along a line parallel to y-axis (x = +a) during this $\vec{F} = - k (y \hat{i} + a \hat{J})$

The first component of force, $- k y \hat{i}$ will not contribute any work because this component is along negative x-direction $(- \hat{i})$ while displacement is in positive y-direction (a,0) to (a,a). The second component of force i.e. $- k a \hat{j}$ will perform negative work

∴ $W_{2} = (- k a \hat{j}) (a \hat{j})$ = $(- k a) (a) = - k a^{2}$

So net work done on the particle W = W₁ + W₂

= $0 + (- k a^{2}) = - k a^{2}$

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If g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass m raised from the surface of earth to a height equal to the radius of the earth R, is

A.

$\frac{1}{2} m g R$

B.

2 mgR

C.

mgR

D.

$\frac{1}{4} m g R$

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Explanation

Gain in potential energy $Δ U = \frac{m g h}{1 + \frac{h}{R}}$

If h = R then $Δ U = \frac{m g R}{1 + \frac{R}{R}} = \frac{1}{2} m g R$

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A lorry and a car moving with the same K.E. are brought to rest by applying the same retarding force, then

A.

Lorry will come to rest in a shorter distance

B.

Car will come to rest in a shorter distance

C.

Both come to rest in a same distance

D.

None of the above

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Explanation

$Stopping distance = \frac{kinetic energy}{retarding force}$ ⇒ $s = \frac{1}{2} \frac{m u^{2}}{F}$

If lorry and car both possess same kinetic energy and retarding force is also equal then both come to rest in the same distance.

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A particle free to move along the x-axis has potential energy given by $U (x) = k [1 - e^{{(- x)}^{2}}]$ for $- \infty \leq x \leq + \infty$ , where k is a positive constant of appropriate dimensions. Then

A.

At point away from the origin, the particle is in unstable equilibrium

B.

For any finite non-zero value of x, there is a force directed away from the origin

C.

If its total mechanical energy is k/2, it has its minimum kinetic energy at the origin

D.

For small displacements from x = 0, the motion is simple harmonic

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Explanation

Potential energy of the particle $U = k (1 - e^{- x^{2}})$

Force on particle $F = \frac{- d U}{d x} = - k [- e^{- x^{2}} \times (- 2 x)]$

F $= - 2 k x e^{- x^{2}} = - 2 k x [1 - x^{2} + \frac{x^{4}}{2!} - .....]$

For small displacement $F = - 2 k x$

⇒ $F (x) \propto - x$ i.e. motion is simple harmonic motion.

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The kinetic energy acquired by a mass m in travelling a certain distance d starting from rest under the action of a constant force is directly proportional to

A.

$\sqrt{m}$

B.

Independent of m

C.

$1 / \sqrt{m}$

D.

m

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Explanation

Kinetic energy acquired by the body

= Force applied on it × Distance covered by the body

K.E. = F × d

If F and d both are same then K.E. acquired by the body will be same

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A body is moving along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to

A.

t^1/2

B.

t^3/4

C.

t^3/2

D.

t²

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Explanation

$P = F v = m a v = m (\frac{d v}{d t}) v$ ⇒ $\frac{P}{m} d t = v d v$

⇒ $\frac{P}{m} \times t = \frac{v^{2}}{2}$ ⇒ $v = {(\frac{2 P}{m})}^{1 / 2} {(t)}^{1 / 2}$

Now $s = \int_{}^{} v d t = \int_{}^{} {(\frac{2 P}{m})}^{1 / 2} t^{1 / 2} d t$

∴ $s = {(\frac{2 P}{m})}^{1 / 2} [\frac{2 t^{3 / 2}}{3}]$ ⇒ $s \propto t^{3 / 2}$

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A particle moves from a point $(- 2 \hat{i} + 5 \hat{j})$ to $(4 \hat{j} + 3 \hat{k})$ when a force of $(4 \hat{i} + 3 \hat{j})$ N is applied. How much work has been done by the force?

A.

8 J

B.

11 J

C.

5 J

D.

2 J

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Explanation

Displacement of the particle,

$∆ \vec{s} = {\vec{r}}_{2} - {\vec{r}}_{1} = 4 \hat{j} + 3 \hat{k} - (- 2 \hat{i} + 5 \hat{j}) = 2 \hat{i} - \hat{j} + 3 \hat{k}$

Force on the particle,

$\vec{F} = 4 \hat{i} + 3 \hat{j} N ∴ Work done, W = \vec{F} . ∆ \vec{s} = (4 \hat{i} + 3 \hat{j}) . (2 \hat{i} - \hat{j} + 3 \hat{k}) = 8 - 3 = 5 J$

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A body of mass 1 kg begins to move under the action of a time dependent force $F = (2 t \hat{i} + 3 t^{2} \hat{j})$ N, where $\hat{i}$ and $\hat{j}$ are unit vectors along X and Y axis, What power will be developed by the force at the time (t) ?

A.

$(2 t^{2} + 4 t^{4}) W$

B.

$(2 t^{3} + 3 t^{4}) W$

C.

$(2 t^{3} + 3 t^{5}) W$

D.

$(2 t + 3 t^{3}) W$

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Explanation

(c) According to question, a body of mass 1kg begins to move under the action of time dependent force,

$F = (2 t \hat{i} + 3 t^{2} \hat{j})$

where $\hat{i}$ and $\hat{j}$ are unit vectors along X and Y axis

$∵ F = m a \Rightarrow a = \frac{F}{m} \Rightarrow a = \frac{(2 t \hat{i} + 3 t^{2} \hat{j})}{1} \{∵ m = 1 k g\} \Rightarrow a = (2 t \hat{i} + 3 t^{2} \hat{j}) m / s$

$∵ a c c e l e r a t i o n, a = \frac{d v}{d t}$

$\Rightarrow d v = a d t$ ...(i)

integrating both sides, we get

$\int d v = \int a d t = \int (2 t \hat{i} + 3 t^{2} \hat{j}) d t v = t^{2} \hat{i} + t^{3} \hat{j}$

$∵$ Power developed by the force at the time t will be given by as

$P = F . v = (2 t \hat{i} + 3 t^{2} \hat{j}) . (t^{2} \hat{i} + t^{3} \hat{j}) = (2 t . t^{2} + 3 t^{2} . t^{3}) P = (2 t^{3} + 3 t^{5}) W$

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