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M P Keshari » PHYSICS – KINEMATICS

PHYSICS – KINEMATICS

This item was filled under [ COMPETITION, NDA, SCRA ]

Physics for Engineering/Medical

Kinematics

Question – 1.

The velocity time graph of a particle is represented by  y = mx + c, the particle is moving with :

(A) Constant velocity                                    (C) Constant acceleration

(B) Variable velocity                                     (D) Constant momentum

Solution : –

We have, the velocity –time graph has positive slope. Hence, the acceleration is constant.

Ans is (C)             [Ans.]

Question – 2.

An aeroplane flying at a constant speed releases a bomb. When the bomb moves away from the aeroplane, it will :

(A) Always be vertically below the aeroplane, if the aeroplane is flying horizontally.

(B) Always be vertically below the aeroplane.

(C) Always be vertically below the aeroplane, if the aeroplane is flying at an angle of 35˚ to the horizon.

(D) Gradually falls behind the aeroplane.

Solution : –

We have, the horizontal component of the velocity of the bomb and the aeroplna are the same. Hence, horizontal displacements would remain the same at all time.            (A)          [Ans.]

Question – 3.

A car travels 150 km east and then 150 km south. Finally, it comes back to the starting point by the shortest route. The speed is 60 kmh–1 throughout the journey. The average velocity is :

(A) 120 kmh–1 (B) 90 kmh–1 (C) 60 kmh–1 (D) 0 kmh–1

Solution : –

We have, the net displacement zero; hence average velocity zero           (D)          [Ans.]

Question – 4.

The x and y coordinates  of a particle at any time t are given by x = 2t + 4t2 and y = 5t. The aceeleration of the particle at t = 5 is given by :

(A) Zero                               (B) 8 ms–2 (C) 16 ms–2 (D) 32 ms–2

Solution : –

We have,             vx =             d/dt(2t + 4t2)     =             2 + 8t

ax =             d/dt(2 + 8t)         =             8

vy =             d/dt(5t)                                =             0

ay =             d/dt(vy)                =             0

The acceleration is along x-axis and is 8 ms–2 (B)                 [Ans.]

Question – 5.

A projectile is thrown at an angle of β with the vertical. It reaches a maximum height H. The time taken to reach the highest point of its path is :

(A) √(H/g)                          (B) √(H/2g)                         (C) √(2H)/g                         (D)√(2H/cosβ)

Solution : –

We have,             H             =             (v2cos2β)/(2g)

Or,                                          vcosβ    =             √(2gH)

And                                        t              =             (vcosβ)/g

=             √(2gH)/g

=             √(2gH)/g              (C)          [Ans.]

Question – 6.

The equation of a projectile is as

y = x√3 – (gx2)/2.

The angle of projection is :

(A) 15˚                                 (B) 30˚                                   (C) 45˚                                   (D) 60˚

Solution : –

We have, by comparing

y = x√3 – (gx2)/2               and        y = x tan θ – (gx2)/(2v2cos2θ),

tan θ      =             √3

Or,                          θ             =             60˚          (D)          [Ans.]

Question – 7.

A particle makes n revolution in one second in a circular orbit of radius r. Its centripetal acceleration would be :

(A) 4πn2r2 (B) 4π2n2r                            (C) 4π2nr                              (D) 4πnr2

Solution : –

We have,             ac            =             rω2

=             r(2πn)2

=             4π2n2r   (B)                          [Ans.]

Question – 8.

A motion in a straight line is supplied with a constant power. The distance travelled in time t would be proportional to :

(A) t3/2 (B) t –3/2 (C) t –1/2 (D) t1/2

Solution : –

We have,             Power (P)            =             (1/2)[(mv2)/t]

Or,                                                          v2 =             (2Pt)/m

Or,                                                          v              =             √[(2Pt)/m]

Or,                                                          dx/dt     =             [√{(2P)/m}]t1/2

Integrating, we get         x              =             [√ {(2P)/m}][(t3/2)(3/2)]

Or,                                                          x              α             t3/2 (A)                          [Ans.]

Question – 9.

Displacement x = t4, the ratio of acceleration and velocity would be :

(A) t/3                                  (B) t/4                                   (C) 3t                                     (D) 3/t

Solution : –

We have,             x              =             t4

Therefore,          velocity                                =             dx/dt                     =             4t3

And                        acceleration       =             d2x/dt2 =             12t2

Therefore,          Acceleraton/Velocity     =             (12t2)/(4t3)

=             3/t          (D)          [Ans.]

Question – 10.

A projectile is projected and it is found that the horizontal range and the maximum height are equal. The angle of projection would be :

(A) θ = tan–1 (4)                (B) θ = tan–1 (3) (C) θ = tan–1 (2) (D) θ = tan–1 (1)

Solution : –

We have,             Horizontal range               =             (u2 sin 2θ)/g

Maximum height             =             (u2 sin2 θ)/(2g)

Both are equal

Therefore,                          (u2 sin2 θ)/(2g)                  =             (u2 sin 2θ)/g

Or,                                                          sin2 θ/2 =             2 sin θ cos θ

Or,                                                          tan θ                      =             4              (A)          [Ans.]

Questiopn – 11.

A constant force directed towards a fixed point is applied to a body in motion. The magnitude of the force varies inversely as the square of the distance from the fixed point. The path of the motion would be :

(A) Parabola                      (B) Hyperbola                   (C) Circle                             (D) Ellipse

Solution : –

We have, the force is directed towards a fixed point and changes the direction of motion without changing its speed. Hence, the path of the motion is a circle.               (C)                          [Ans.]

Question – 12.

A particle performing uniform circular motion with constant speed v in a circle of radius r. The tangential acceleration would be :

(A) Zero                               (B) v r                                    (C) v2 r                                  (D) v2/r

Solution : –

We have, a particle performing uniform circular motion has radial acceleration only and no tangential acceleration.                                (A)                          [Ans.]

Question – 13.

The ratio of time of ascent to the time of descent of a projectile projected at an angle of 50˚ with the horizontal is :

(A) 1 : 2                                (B) 2 : 3                                 (C) 3 : 2                                 (D) 1 : 1

Solution:  –

We have, in a projectile,

Time of ascent                  =             Time of descent

=             (u sin θ)/g

Therefore,          Time of ascent/Time of descent                                =             [(u sin θ)/g]/[(u sin θ)/g]

=             1 : 1        (D)          [Ans.]

Question – 14.

A ball of mass m performs uniform circular motion. The radius of circular path is R. The linear momentum is represented by P. The radial force acting on the particle would be :

(A) P2mR                             (B) P2m/R                            (C) P2/mR                            (D) P2R/m

Solution : –

We have,             Momentum, P =             mv          —           —           —           (1)

Radial force, F    =             mv2/R   —           —           —           (2)

Putting, v             =             P/m in (2), we get

F              =             (m/R)(P/m)2

=             P2/mR   (C)          [Ans.]

Question – 15.

A particle excutes circular motion under a central attractive force inversely proportional to distance R. The speed of the particle would be :

(A) Dependent on R                                       (C) Dependent on R2

(B) Independent on R                                   (D) Independent on 1/R

Solution : –

We have,             Centripetal force, F         =             mv2/R

Therefore,                                                          F              α             1/R

Therefore,          v does not depend on R.              (B)          [Ans.]

Question – 16.

A particle of mass m excutes a circular motion in a circular path of radius R. The centripetal acceleration varies as square of time given by βRt2. The force acting on the particle provides a power given by :

(A) mβR2t           (B) mβR2t2 (C) mβRt2 (D) mβrt

Solution : –

We have, tangential acceleration exists, hence power is required.

Acceleration       =             v2/R       =             βRt2

Or,                          v              =             Rt√β

Tangential acceleration                                 =             aT =             dv/dt

Or,                          aT =             d(Rt√β)/dt

=             R√β

Force     =             maT =             mR√β

Power   =             force×velocity

=             (mR√β)×(Rt√β)

=             mβR2t                   (A)                          [Ans.]

Question – 17.

A projectile is projected at an angle θ with horizontal. When the particle makes an angle β with the horizontal, its speed v is given by :

(A) v = u cos θ cos β        (B) u cos θ sin β                                (C) u sin θ cos β                                (D) u cos θ sec β

Solution : –

We have, as horizontal component of velocity remains constant.

Therefore,          u cos θ                  =             v cos β

Or,                                          v              =             u cos θ sec β       (D)          [Ans.]

Question – 18.

A projectile is projected from a point and the point is choosen as origin. In t seconds, vertical distance y = 8t – 15 t3 and horizontal distance x = 6t – 13t3. The velocity of projection of projectile is given by :

(A) 8 m/s                             (B) 10 m/s                           (C) 12 m/s                           (D) 14 m/s.

Solution :

We have,             y              =             8t – 15 t3

dy/dt     =             8 – 45 t2

At,          t              =            0,            vy =             8

x              =             6t – 13 t3

dx/dt     =             6 – 39 t2

At,          t              =             0,            vx =             6

(Velocity)2 =             vy2 + vx2

=             82 =             62

=             100

Velocity                =             10.          (B)          [Ans.]

Question – 19.

A particle describes parabolic paths in time t1 and t2 and covers the same horizontal range R. Then t1t2 is proportional to :

(A) R3 (B) R2 (C) R                      (D) √R

Solution : –

We have,             Time of flight                     =             (2u sin θ/g

Horizontal range               =             (u2 sin 2θ)/g

As, the range is the same, angle of projection should be θ and 90˚ – θ.

Therefore,                          t1 =             (2u sin θ)/g

And                                        t2 =             {2u sin (90˚ – θ)}/g

=             (2u cos θ)/g

Therefore,                          t1t2 =             [(2u sin θ)/g]×[(2u cos θ)/g]

=             (4u2 sin θ cos θ)/g2

=             2[(u2 sin 2θ)/g]/g

=             2R/g

Hence,                                  t1t2 α             R             (C)          [Ans.]

Question – 20.

If a car covers 2/5 of the total distance with speed v1 and 3/5 distance with speed v2 then the average speed is :

(A) (v1 + v2)/2                    (C) 5v1v2/(3v1 + 2v2)        (C)2v1v2/(v1 + v2)              (D) v1v2

Solution : –

We have,             Speed                   =             Distance/Time

Time                      =             Distance/Speed

Let total distance             =             S,

Total time            =             t1 +             t2 =            (2S/5)/v1+(3S/5)/v2

Therefore,          Average velocity              =             S/{(2S/5)/v1+(3S/5)/v2}

=             5v1v2/(3v1 + 2v2)               (B)          [Ans.]

Question – 21.

The acceleration of a particle is increasing linearly with time t as at. The particle starts from the origin with an initial speed of u. The distance travelled by the particle in time t would be :

(A) ut + 1/6 at3 (B) ut + 1/4 at3 (C) ut + 1/3 at3 (D) ut + 1/2 at2

Solution : –

We have,             acceleration       =             dv/dt     =             at

Therefore,                                          dv           =             atdt

Therefore,                                          uvdv     =             0tatdt

Or,                                                          v – u      =             at2/2

Or,                                                          v              =             u             +             at2/2      =             ds/dt

Or,                                                          ds           =             udt         +             (at2/2)dt

Integrating, we get                         s              =             ut            +             1/6 at3 (A)          [Ans.]

Question – 22.

A particle moves in a straight line so that its displacement x in meter at time t in second is given by x2 = 1 + t2. The acceleration  of the particle in m/sec2 at time t is given by :

(A) t/x2 (B) 1/x – t/x2 (C) 1/x – t2/x3 (D) t2/x2

Solution : –

We have,             x2 =             1              +             t2

Or,                                          x              =             (1 + x2)1/2

Therefore,                          dx/dt     =             (1/2) × (1 + t2)–1/2 × 2t

=             t(1 + t2)–1/2

Acceleration       =             d2x/dt2 =             t × (–1/2) × (1 + t2)–3/2× 2t + (1 + t2)–1/2

=             1/x – t2/x3 (C)          [Ans.]

Question – 23.

A projectile is projected with an initial speed of (xi + yj ) m/sec. If the range of the projectile is double the maximum height reached by it then :

(A) x = y                               (B) y = 2x                             (C) y = 3x                             (D) y = 4x.

Solution : –

We have,             u sin θ   =             y             and       u cos θ =             x.

Therefore,                          H             =             (u2 sin2 θ)/2g      =             y2/2g

And                                        R             =             [u2(2 sin θ cos θ)]/g          =             2xy/g

It is given that                    R             =             2H

Therefore,                          2xy/g     =             2y2/2g

Or,                                          y              =             2x           (B)          [Ans.]

Question – 24.

A stone tied to the end a string of 1 m long is whirled in a horizontal plane in a circle with a constant speed. The stone makes 30 revolution in 1 minute. The magnitude and direction of acceleration of the stone would be :

(A) π2 ms–2 along the radius toward the centre

(B) π2 ms–2 along the tangent to the circle

(C) π2 ms–2 along the radius away from the centre

(D) π2 ms–3 away from the centre.

Solution : –

We have,             a              =             rω2;        ω             =             2πν

Again                     30 revolution     =             60 sec

Or,                          1 revolution        =             2 sec

Therefore,                          ν              =             1/2 Hz

And                                        a              =             rω2

=             1×(4π2)/4

=             π2 ms–2 (A) & (C)              [Ans.]

Question – 25.

The speed of a boat in still water is u and that in river is v. The time taken by the boat to reach a place downstream at a distance d and come back to the original place is :

(A) ud/(u2 – v2)                (B) 2ud/(u2 – v2)               (C) 2vd/(u2 – v2)               (D) vd/(u2 – v2)

Solution : –

We have, the velocity downstream = u + v and velocity upstream = u – v. (u > v), otherwise boat will not return back upstream.

Therefore,          Time for to and fro trip                  =             d/(u + v) + d/(u – v)

=             2ud/(u2 – v2)      (B)          [Ans.]

 

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