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8401.

Two sources of emf 6V and internal resistance 3Ω and 2Ω are connected to an externalresistance Ras shown. If potential difference across source A is zero, then value of R is7.(r

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8402.

33. A ball is thrown at a speed of 40 m/s at an angle of 60°with the horizontal. Find (a) the maximum heightreached and (b) the range of the ball. Take g 10 m/s2.

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8403.

A bullet is fired from a gun with a speed of 100ms in a direction of 30° abovethe horizontal then calculate i) maximum height reached ii) Time of flightand iii) horizonatl range of bullet. Given g =9.8 ms 2.

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H = u^2 sin2 theta/g put all the values you will get it

8404.

33. A ball is thrown at a speed of 40 m/s at an angle of 60with the horizontal. Find (a) the maximum heightreached and (b) the range of the ball. Take g 10 m/s .

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8405.

EDPROBLEMSt an angle 60 withthe horizontal. What is the maximum height reached by t(Take g # 9.8 ms)m 1: A body is projected with a velocity of 49 ms1 at an athe body So

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8406.

18. If the initial velocity of a projectile be doubled. Keeping the angle of projection same, the maximum heightreached by it will(A) Remain the same(8) Be doubled(C) Be quadrupled (D) Be halvesd

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8407.

e: 90 MinTime: 9oA particle moves in a circle of radius 5 cmwith constant speed and time period 0.2π s.The acceleration of the particle is1.(a) 15 m/s(c) 36 m/s2Which of the following :(b) 25 m/s2-(d) 5 m/s2.

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8408.

why does the temperature remain constant during the change of state?

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8409.

31.The angle between velocity vector and acceleration vector in uniform circular motion is(2) 180°(4) 45(3) 90

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When the object is moving inuniform circular motion, then theangle betweenthevelocity vectorandacceleration vectoris equal to 90 degrees.

In this case, the type of acceleration caused is generally called ascentripetal acceleration.

Thus acceleration vector forms 90 degree angle with the velocity vector.Hence, option (3) is correct.

8410.

3. The displacement of a particle, moving in a straight line,is given by s 2r +21 +4 where s is in metres and t inseconds. The acceleration of the particle is(a) 2 m/s2(c) 6 m/s2(b) 4 m/s2(d) 8 m/s2

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8411.

3.A body starts from rest. What is the ratio of thedistance travelled by the body during the 4th and3rd second:(17s(4) 3/7(2) 5/7(3) 7B

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8412.

51. A block is moving along y-axis with velocity A 4 on a plank relative to ground and the plank is moving along x-awith velocity Vp 3i at any instant of time. The unit vector in direction of friction force acting on block at this momeof time is3i+ 4j53i -4j-3i + 4j(3) 5(4)1-3i-4j5

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velocity of block w.r.t plank = Vb-Vp = 4j-3i and unit vector will be = (-3i+4j)/5

now , friction always acts opposite to the velocity of the body

so, unit vector along the Frictional force = -[-3i+4j]/5 = (3i-4j)/5

option B

8413.

30. Does the linear velocity of a planet remain constant in its orbit?

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In order to keep the arealvelocity constant, thelinear velocityof theplanetgoes on changing. ... The square of the time period of revolution of aplanetaround the sunisdirectly proportional to the cube of the semi major axis ofitsellipticalorbit.

8414.

plane.ThereA condacting rod of length I is hinged at point O It is free to rotate in a verticalexistsa uniform magnetic field ธิ in horizontal direction. The rod is released from the positionshown Thepotential difference between the two ends of the rod is proportional to(C) sin e(d) (sin e)1/2

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8415.

A body starts from rest from the origin with anacceleration of 6m/s2 along the x-axis and8m/s2 along the y-axis. Its distance from theorigin after 4 seconds will be:(1) 56 m(3) 80 m95.(2) 64 m(4) 128 mThe x and y coordinates of a particle at any timet are given by x 7t + 4t2 and y 5t, where x96.and y are in metre and t in seconds. Theacceleration of particle at t 5 s is:(1) Zero(3) 20 m/s2(2) 8 m/s2(4) 40 m/s2

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Using Newton's second formula of motion

Sx = 1/2*ax*t² = 1/2*6*4² = 48m

and Sy = 1/2*ay*t² = 1/2*8*4² = 64m

so, distance from origin is = √48²+64² = 80m

option 3.

acceleration in x direction = d²x/dt² = d(7+8t)/dx = 8m/s²

acceleration in y direction = d²y/dt² = d(5)/dt = 0

so only acceleration is 8m/s².

option 2

Can u answer it using projectile motion.

answer will be same im any case...

for projectile eliminate t from both the equation.. and write equation of y in terms of x.

8416.

If x - at + bt2, where x is the distance travelled by thebody in kilometres while t is the time in seconds, then theunits of b are(a) km/s(b) km-s(d) km-s2(c) km/s2

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X = at + bt^2

Units of at and bt^2 should be km. Then it is possible to get “x” in km.

Therefore,km = b(s)^2b = km/s^2

Units of b is km/s^2

8417.

31.For a body traveling with uniform acceleration, its final velocity is v- V180 - 7x,where x is the distance traveled by the body. Then acceleration isA) -3.5m/s2 B) 3m/s2 C)-2.5m/s2 D ) 2.5m/s2

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8418.

8. A particle move along the parabolic pax y2 + 2y + 2 in such a way thaty-component of velocity vector remain 5 m/s duthe motion. The magnitude of the accelerationthe particle is(1) 50 m/s2(2) 100 m/s2(3) 10/2 m/s2(4) 0.1 m/s2

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8419.

equalA drop of liquid of mass 2g falling from height 2km.It hits the ground with speed 25 m/s. What is thework done by (i) the gravitational force? (ii) thenknown force? (g = 10m/s^2)

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8420.

A smooth ring of mass m can slide on a fixed horizontal rod. A string tied to the ring passes overpulley B and carries a block C of mass 2m as shown below. as the ring starts sliding.2g cos1+2 cos2g(A) The acceleration of the ring is(B) The acceleration of the block isC) The tension in the string is1+2cos2mg1+2cos2 8

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8421.

A siphon in use is demonstrated in the following .The density of the liquid flowing insiphon is 1.-5gm/ ce . The pressure difference between the point P and S will be42.1020 cm1) 10sN/m2) 2x1Ợ N/ m3) zero4) Infinity

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The answer is option C.

8422.

In the dimensional analysis of the equation, (velocity)comes out to be 3.(pressure difference)s2 x (density)32. The value of

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8423.

A siphon in use is demonstrated in the following The density of the liquid flowing insiphon is 1.5gm/cc ·The pressure difference between the point P and S will be42.10 cm20 cm,SI1) 10'NIm2) 2x10'N/m3) zero4) Infinity

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8424.

Pressure 3 m below the free surface of a liquid is 15KN/m2 in excess of atmosphere pressureDetermine its density and specific gravity. [g 10 m/sec]nectad to a samo tube as shown in fiouce netermine difference of

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8425.

1. What is Projectile motion?

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A projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity.

8426.

The density of vanadium in S:96alem S. convert the density inS units of Bg/m3

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the answer of this question is 5960 kg/m3

answer=5960kg/m3is a write answer

8427.

20. For the zero acceleration of the m, as shown in figure, coefficient of friction between m2 and the horizontal surfaceis μ. The minimum value of m is(a) m2(b) m(c)+ m2μ+ m

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8428.

Dimensional analysis of the relation (Energy)"(Pressure difference)s (Volume)32 gives thevalue of n as

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8429.

Two Body of masses 1kg & 2kg having coordinate(i,1) & (2,2), find com?in A

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8430.

Q6 Two Body of masses 1kg & 2kg having coordinate(1,1) & (2,2), find com?

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body A has mass of 1 kg and location (1,1)

body B has mass of 2kg and location (2,2)

mcxc= m1x1+ m2x2

(1+2) xc= (1*1) + (2*2)

xc= 5/3

similarly

mcyc= m1y1+ m2y2

(1+2)yc= (1*1)+(2*2)

yc= 5/3

hence the co-ordinates of the center of mass are ( 5/3 ,5/3)

8431.

How can the number of neutrons in an atom be determined?

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no. of neutron can be determined by

(Atomic mass - atomic no. of that atom(z)) no. of neutron = A-Z

8432.

7.- Find dimension of a × b in relation.2F =,HfFis force, x is position and t is time.bt

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8433.

Write the dimensions of a/b in the relation PPis pressure, s is distance and t is timeThe disniacement of a hondy is given to be proportional to the cube of time elapsed What is the

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8434.

EQUATION FOR VELOCITY-TIMERELATION

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8435.

8.5.2 EQUATION FOR POSITION-TIMERELATION

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8436.

4.1.a .)Derive graphically the equation for velocity - time relation

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8437.

A boat is sent across a river with a velocityof 8 kmfh. Ifboat is 10 km/h, the river is flowing with avelocity of36.the resultant velocity of the

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The boat’s velocity and the river velocity will give the boat a new resultant velocity.

As the boat’s velocity is perpendicular to river velocity,

Assume that,

VB= Velocity of boat

VR= Velocity of river

Apply pythagoras theoram,

Resultant velocity2= VB2+VR2

102=82+VR2

100 = 64 +VR2

VR2 = 36

VR = 6 km/h

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8438.

1.Displacements is related with time t in theequation s = 5+2 4+ + 3. The acceleration isequal to

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time =distance by speed apply this formula and get the answer

time equal to distance by speed

double differentation of displacement gives accerlation by doing diffrentiation we will velocity v=5(2t)+4(1)+0;v=10t+4 again doing differention we get a = 10(1)+0; a=10

acceleration is double derivative of displacement function so resultvelocity = 10t + 4(derivative of displacement) therefore acceleration = 10

8439.

23. A boat is sent across a river wi8 km/hr. If the resultant velocity of boatthen velocity of the river is(1) 10 km/hr(3) 6 km/hr(2) 8 km/hr(4) 4 km/hr

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8440.

2. A bus decreases its speed JUIL80 km h-' to 60 km h in 5 s.Find the acceleration of the bus.

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8441.

Conservation of linear momentum is equivalentto-

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Newtons First law of motion.

8442.

A ball whose mass is 100 g is dropped from a heightof 2 m from the floor. It rebounds vertically up-wards after colliding with the floor to a height 1.5m. Find (a) the momentum of the ball before andafter colliding with the floor, (b) the average forceexerted by the floor on the ball. Assume that colli-sion lasts for 10^- 8 s.

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V2−U2 = 2gS Here given that, U = 0 m/s, S = 1.5 m. So,V2−02 = 2×10×2 ⇒V=40−−√ m/s.So, momentum of the ball is,Pi=mV = 0.1×40−−√ = 0.1×6.325 = 0.633⇒ Pi=0.633 kg.m/s

After collision, it is given that the ball rises to a height S' = 1.5 m.So, velocity with which the body rises is,

V2−U2=−2gS' ⇒ 02 − V'2 = −2×10×1.5 ⇒V' = 30−−√ m/s.So, the momentum of the ball, when the ball is bouncing back.

Pf=mV' = 0.1×30−−√⇒Pf=0.547 kg m/s.

Therefore, the average force exerted by the floor on the ball is,

Favg=∆PDt = 0.633−0.54710−8=0.086×108 ⇒Favg=8.6×10^6N

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8443.

A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. the work done by the forceof gravity during the time the particle goes up isAIEEE 2006, 1.5/180,-1]

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M = 0.1 kgV1 = 5 m/sV2 = v1 - g*t = 5 - 9.8 * t = 0 (highest point)t = 0.51 sec

S = v1 * t - 1/2gt^2 = 5 * 0.51 - 4.9 * (0.51)^2 = 1.276 m = h

W = m * g * h = 0.1 kg * 9.8 m/s^2 * 1.276 m = 1.25 joulesor alternatively:W = initial Ek - final Ek = mgh = 1/2 m * V1^2 = 1.25 joules(final kinetic energy = 0, because of the work done by gravity, all Ek is transformed into potential energy Ep = mgh)

8444.

2.State and derive law of conservation of linearmomentum in case of colliding bodies.

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8445.

11. Which substance can possess three states of matter easily12, Air can ha

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Mercury and water are not the only substances capable of existing in three distinct states of matter. In fact, all of the elements, of which mercury is one, may exist insolid,liquid, or gas forms.

8446.

A bus decreases its speed from 80 km h-1 to 60 km h"1 in 5 s. Find the acceleration of the bus.

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8447.

Drive all the three equation of motion by graphical method

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,

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8448.

Derive the velocity- time equation of motion by Graphical method

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8449.

Define the following terms :-(a) Newton's frist law of motion(c) Newton's third law of motion(e) Inertia of rest

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1.Newton's First Lawstates that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. It may be seen as a statement about inertia, that objects will remain in their state of motion unless a force acts to change the motion.

2.A force is a push or a pull that acts upon an object as a results of its interaction with another object. These two forces are called action and reaction forces and are the subject ofNewton's third law of motion. Formally stated,Newton's third lawis: For every action, there is an equal and opposite reaction.

3.The tendency of a body atrestto remain atrest, or of a body in motion to continue moving in a straight line at a constant speed unless a force is applied to it. Mass is a measure of a body'sinertia.

8450.

For a hydrogen atom, which electronic transition would result in the emission of a photon with the highenergy?(1) 2s- 3p(2) 3p- 6d(3) 4p-2s(4) 5f-3d

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