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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 13051. |
The force of gravitation is(a) Repulsivee) conservative(b) Electrostatics(d) Non conservativeiulum on a freely moving artifcial satellite is [AFNIC 2002) |
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Answer» Conservative.Because the work done by gravity doesn't depend on the path taken, we call gravity a conservative force. The force exerted by a spring is another example of a conservative force. |
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| 13052. |
ĐĄĐThe minimum power of eye lens is 40 D. If the far point of normal eye is infinity, find the size of the eye ball.CI |
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Answer» Size of the eyeball would indicate the distance between the eye lens and the retina. As the image is formed on the retina, the size of the eyeball would be equal to the image distance (v).Now, as the object distance is infinite, we know that the image will be formed on the focus. So, here v = f (focal length). Now, as we know that,P=1/fP= 40 D40=1/ff=140f= 0.025 meters = 2.5 cmfocal length=2.5 cm Therefore, size of eye ball is 2.5 cm |
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| 13053. |
29.At the topmost point of its path, a projectile has acceleration of magnitude2) g922 |
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Answer» option (2) is correct. acceleration due to gravity is constant on the object thoughout its flight. So the acceleration of the projectile is equal to the acceleration due to gravity,9.81 |
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| 13054. |
E X E R C I S E S1.A solar water heater cannot be used to get hot water on(a)(c)a sunny day.a hot day(b) a cloudy day(d) a windy day |
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Answer» b is the right answer a cloudy day .......... |
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| 13055. |
A distance of 2.00 mm separates two objects of equal mass.If the gravitational force between them is 0.0104 N, find themass of each object.6) |
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Answer» Given the distance r = 2/1000 m, the force between them F =0.0104 N, the mass of the two object can be calculated using formula: F = G(m1m2)/r^2 since the mass are equal F = G (m^2)/r^2 And where G = is the gravitational constant (6.67E-11 m3 s-2kg-1) The mass of the two objects are 24.96 kg |
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| 13056. |
p block contains(1) Only metals(2) Only non metais(3) Only metals and non metals(4) Metals, non-metals and metalloids |
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Answer» Thep-blockis home to the biggest variety of elements and is the only blockthatcontainsall three types of elements: metals, nonmetals, and metalloids. Generally, thep-block elements are best described in terms of element type or column number. aap math se ho kya |
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| 13057. |
Calculate the graviculatethegravitahona e betueen he earthоо9istancence, betweenthem35% 168 n |
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| 13058. |
pecontact.rive an expression for eqvivalent focal length of two thin convex lenses inL1 L2I1O,V1'ぐ |
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Answer» tq... |
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| 13059. |
(3 21 रथ 2 .(42224. हक: A ub ,n‘# Jends&Lk o pehkyy KAy ०which M1 fance Peison 7 wi |
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Answer» A myopic person has clear vision when looking at objects close to them, but distant objects will appear blurred. Like my answer if you find it useful! |
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| 13060. |
orObtain the formula of torque acting on an electric dipolein a uniform electric field. Two point charges of +1 x10-6coulomb and-1Ă10-6 coulomb are situated at a distance of2.0 em. The electric dipole is placed in a uniform electric fieldof 1x10s volt/metre. Find the potential energy of the dipoleI Ans. -2 x10-3J ]5in stable equilibrium.7 Th |
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| 13061. |
Q.11) A force F, acts on a particle so as to accelerateit from rest to a velocity v. The force F, is thenreplaced by F which decelerates it to rest.(1) F1 must be equal to F2(2) F, may be equal to F2(3) F, must be unequal to F2(4) of these |
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Answer» ans is (4) none of these Option 1 is the correct answer. As force F2 is able to decelerate F1 so it must be equal to it direction of force will be change to decelerate so 1st cant be the anwer |
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| 13062. |
force F is needed to break a copper wire having radius R. The force needed to break a copper wire ofradius 2 R will be(A) F/21. A(B) 2 F(C) 4 F(D) F/4 |
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| 13063. |
Diatomic gas at pressure 'P' and volume 'Vis compressed adiabatically to 1/32 times the forginal volume. Then the final pressure is |
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| 13064. |
v. Observe the following diagram and answer the questions.i mim2I) Which law do we understand from the above diagram? State the lawii) State the mathematical equation for the law.iii) How will the value of force F change if the mass m, is increased to |
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Answer» Newton's lawof universal gravitation states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. 2) F= m1m2/r^2 |
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| 13065. |
i. An electric heater of 500 W operates 12hr/days. What is the cost of the energy to operate it for 30 daysat 4 rupees per kWh?ii State loules laww of Heating |
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Answer» Load=500 watt. Running hours per day= 12 hours Daily consumption= 12 X 500= 6000 watt hour or 6 Kwh Rate per unit= 4 Rs. Daily Cost = 6 X 4 = 24 Rs. Monthly Cost= 24 X 30 = 720 Rs |
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| 13066. |
In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1kW. The volfage of the electric mains is 220 V. The minimum capacity of the main fuse of thebuilding will be(A) 8 A(C) 12 A(B) 10 A(D) 14 A |
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| 13067. |
5) Observe the following fgure and complete the table10W work on 220 varallel, how much02Fd/Pointson, electric heater,(i) Position of the object(ia) Position of the imageii) Size of the imagev) Nature of the imageis used, not pure490 m. How much8 m/s)P.TO. |
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Answer» between F and lensbetween F and lensLarger than the objectErect Virtual |
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| 13068. |
48. An electric heater marked 550 W is used for8 hours a day. If the rate of electricity charge is Rs. 0.20per unit, then the total expenditure would be :(a) Rs.0.38 0 (6) Rs. 7.70(c) Rs.0.88(d) Rs. 188119oo |
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Answer» rupees 7.70 is there total expenditure solve kr k btao I don't understand what you siad Option c is correct as:- |
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| 13069. |
3. A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hoursof machine time and 3 hours of craftman's time in its making while a cricket battakes 3 hour of machine time and 1 hour of craftman's time. In a day, the factoryhas the availability of not more than 42 hours of machine time and 24 hours ofcraftsman's time.(6) What number of rackets and bats must be made if the factory is to workat full capacity?(ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, findthe maximum profit of the factory when it works at full capacity. |
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| 13070. |
busservicewithabusleayinginTwo towns A and B are connected by a regulareither direction every T minutes. A man cycling with a speed of 20 km h-1 in thedirection A to B notices that a bus goes past him every 18 min in the direction ofhis motion, and every 6 min in the opposite direction. What is the period Tof thebus service and with what speed (assumed constant) do the buses ply on theroad? |
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| 13071. |
4.eton of velocity at the hghest point of a projectnstif it is movingparrot flying at a speed 5m/s crosses a train. 100m long within 10s. Calculate the train's specd it(a) in the same direction (b) opposite direction. (3)5, Show that fiudes of either vector. (3) |
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| 13072. |
ofabodymovingalongx-axıs,213. If v is the velocitythen acceleration of body is\begin{array}{ll}{\frac{d v}{d x}} & {(2)} & {-v \frac{d v}{d x}} \\ {{ } & {\text { (4) } v \frac{d x}{d v}}\end{array} |
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Answer» Option (2) is correct. |
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| 13073. |
Two towns A and B are connected by a regular bus service with a bus leaving ineither direction every Tminutes. A man cycling with a speed of 20 km h in thedirection A to B notices that a bus goes past him every 18 min in the direction ofhis motion, and every 6 min in the opposite direction. What is the period T of thebus service and with what speed (assumed constant) do the buses ply on theroad? |
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Answer» thanks amazing loving it |
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| 13074. |
A particle moves along an arc of a circle of radiusR. Its velocity depends on the distance covered asv-av/s , where a is a constant then the angle Îąbetween the vector of the total acceleration andthe vector of velocity as a function of s will be(A) tana #-(B) tana2s / R2R(C) tan Îą(D) tano2R |
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| 13075. |
Q.2. Find sibQ3. Draw displacement-time graph for a body moving along positive x-axis with increasingvelocity and starting point is taken as origin. |
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| 13076. |
The position x of a particle of mass 2 kg dependson time t (in second) as x(4 +3f-t+ 5) mThe magnitude of force acting on the particleatt=Isis(1) 20 N(2) 30 N(3) 60(4) 90 N |
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Answer» we know that d^2x/dt^2= acceralation where x is displacementso Acceralation = first time diffrentiation= 12t^2+6t-1+0second time = 24t+6t= 130F= ma2*30= 60NAnswerHIT THE LIKE BUTTON IF YOU ARE SATISFIED |
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| 13077. |
2.A plane mirror is on y-z plane facing positive x-axis. A point object is present at (10,5). The mirroris translated by 2 units along positive x and 3 units along positive y-direction. Final image of thepoint object is at1) (-10,-5)2) (-8, -3)3) (-6,5)4) (-6,-2)flot is |
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Answer» 3) is the correct answer |
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| 13078. |
2. A farmer moves along thedeboundary of a square field of st10 m in 40 s. What will be themagnitude of displacement of thefarmer at the end of 2 minutes 20seconds from his initial position? |
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| 13079. |
A bullock is pulling a cart. The cartes. There is a force on the cart and themovcart has moved. Do you think that work isdone in this situation? |
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Answer» yes, work is done in such a situation.There is a force applied on the cart by the bullock and the cart is displaced . Therefore,work is done in such a situation |
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| 13080. |
A mass m is moving with a constant velocityalong a line parallel to the x-axis away fromthe origin. Its angular momentum withrespect to the origin :(a) is zero(b) remains constant(c) goes on increasing (d) goes on decreasing. |
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Answer» remains constant b) remains constant the anglular momentum with respect to origin always constant |
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| 13081. |
15. A positive point charge O is fixed at the origin. A dipolewith dipole moment p is placed along X-axis far awayfrom the origin with p pointing along positive X-axis.Calculateo ineegy af the dipol when i eacries a disance dfrom the origin.(ii) the force experienced by the charge q at this moment. |
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| 13082. |
A point charge of +8μC cxperiences a force of 2000 N at a point, what is the electric fieldintensity at that point ?a ioC ura tnaratod by a distance of 20 cmm. What is the E at a point |
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Answer» The intensity at the given point will be E=F/ q ; So from the given formula the intensity of the given point will be E= 2000/8*10^-6; Which is equal to E= 2.5 *10^8 Newton/coulomb; |
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| 13083. |
Moment of a force of magnitude 20 N acting along positive x direction at point (3m, 0, 0) about the point (0, 2,of magnitude 20 N acting along positive x direction at point(3m, 0,0) about the point (0, 2,C(in Nm) is(1) 20(3) 40(2) 60(4) 30 |
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| 13084. |
A particle starts from the origin, goes alongthe x - axis to the point (20 m, 0) and thenreturns along the same line to the point (-20m, 0). The distance and displacement of theparticle during the trip are1) 60m, -20m 2) 40m, -40m3) 40m, 40m4) 80m, 40m |
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| 13085. |
An athlete completes a circular path of radius 'r'in40s. Find his displacement at the end of2 minutes 20seconds9.What dThe can |
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Answer» Radius is r. Circumference is = 2 pi r. Time is 40 s. Every 40 seconds he comes back to original position l. So in 140 s he would cover 3 and half circles. Hence Total distance is 2r pi x 3.5 = 7r pi. Total displacement will be = 2r since ball will be directly across the diameter |
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| 13086. |
Pressure cannot be measured in(a) Nm2(b) Bar(c) Pa(d) Kg. wt. |
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Answer» Pressure cannot be measured in Ans:(d)Kg.wt... |
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| 13087. |
17. A charged particle is projected with velocity vo along positivex-axis. The magnetic field B is directed along negative z-axis between x = 0 and x = L. The particle emerges out (atx L) at an angle of 60° with the direction of projectionFind the velocity with which the same particle is projected(at x 0) along positive x-axis so that when it emerges out(at x = L), the angle made by it is 30° with the direction ofprojection(1) 2Vo(3) vo /3(2) V/2(4) vo3 |
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| 13088. |
m +m2(0.015+1.2)of mass 60 kg jumps with a horizontal velocity of 6 m s1 onto a stationary cartEXAMPLE 2.28: Awith frictionAssume noless wheels. The mass of the cart is 6 kg. What is her velocity as the cart starts movinexternal unbalanced force is acting in horizontal direction.)CCE 2013]SOLUTION: Here, mass of the girl, m1 = 60 kg: initial horizontal velocity of the girl, ul-6 m s-1; mass of the6 kg and initial velocity of the cart, u: O |
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Answer» Similar problem. Just replace 40 by 60and 5 by 6 no external force => momentum is constant. momentum of girl = 40 * 5 = 200 kg-m/sec momentum of girl and cart = total momentum of the girl + momentum of cart = 200 + 0 = 200 kg m/sec speed = 200 / (40 + 3) = 4.65 m/sec4.1 thanks |
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| 13089. |
A particle starts from the origin goes along the axis x-axis to the point (20m,0) and then return along with the same line to the point (-20m,0). find the distance and displacement of the particle during the trip |
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| 13090. |
2. A particle starts from the origin, goes along the X-axisto the point (20 m, 0) and then returns along the sameline to the point (-20 m, 0). Find the distance anddisplacement of the particle during the trip. |
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| 13091. |
ispiatener2. A particle starts from the origin, goes along the X-axisto the point (20 m, 0) and then returns along the sameline to the point (-20 m, 0). Find the distance anddisplacement of the particle during the trip.-5.0 |
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| 13092. |
A particle revolving in a circular path completes first one third of circumference in 2 s, while nextone third in 1 s. Calculate the average angular velocity of particle. (in rad/s) |
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| 13093. |
Two plane mirrors are placed at an angle α so that a ray paral-lel to one mirror gets reflected parallel to the second mirrorafter two consecutive reflections. The value of a will be(a) 30°(b) 60 (c) 75° (d) 90 |
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| 13094. |
Given a point source of light, which of thefollowing can produce a parallel beam of light ?(a) Concave lensTwo plane mirrors inclined at 90° to each 1other(c) Convex mirror(d) Concave mirror |
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Answer» option D is correct |
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| 13095. |
Calculate the gravitational force of earth acting on a person of mass 60 kg. (Given mass of earth-6102kgand radius of earth-6.4x 106 m, G- 6.7x 10-11 Nm2 kg-2) |
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| 13096. |
WWaE Wll 8CHs velocits 3 s after the start ?) |
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Answer» जब एक लटकते हुए कालीन को छड़ी से पीटा जाता है, तो उसमें से धूल के कण हटा दिए जाते हैं क्योंकि धड़कन पर कालीन अचानक आगे बढ़ जाता है और बाकी की जड़ता के कारण धूल के कण आराम करते रहते हैं। इसलिए धूल के कण नीचे गिर जाते हैं और कालीन साफ हो जाता है। |
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| 13097. |
what is uniform circular motion? |
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Answer» Uniform circular motioncan be described as themotionof an object in acircleat a constant speed. As an object moves in acircle, it is constantly changing its direction. At all instances, the object is moving tangent to thecircle. Uniform circular motion defines the motion of an object traveling at a constant speed around a fixed center point or axis. The object travels around a curved path and maintains a constant radial distance from the center point at any given time. Realistically speaking, a perfect circle does not exist, but it is useful to study the case of a perfect circle in order to understand how an object might move around an ellipse and to approximate the motion of an object that is almost circular in nature. Some examples of this type of motion are the orbit of a planet, a car going around a circular track or a conical pendulum. The motion of the particle along the circumference of the circle , with constant angular velocity is known as uniform circular motion. The motion of a body in a circle with a constant speed is known as uniform circular motion |
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| 13098. |
Differentiate between non-uniform and uniform circular motion |
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| 13099. |
A rotating table completes one rotation in 10 sec.and its moment of inertia is 100 kg-m2. A personof 50 kg. mass stands at the centre of the rotatingtable. If the person moves 2m from the centrethe angular velocity of the rotating table(in rad/sec). will be:2π(1)30 (2) 20m30 (3) 23 (4) 2T |
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Answer» Net torqe about the centre of rotation is zero hence angular monetum is conserved here. Therfore, I1w1 = I2w2 100*(2pi/10) =(100 + 50*4)*w 20pi = 300w2 w2 = 2pi/30 |
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| 13100. |
(2)Uniform circular motion (U.C.M.) is motion ofparticle along circumference of circle withconstant linear speed (or constant angular speed) |
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Answer» Uniform circular motion can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. At all instances, the object is moving tangent to the circle. Since the direction of the velocity vector is the same as the direction of the object's motion, the velocity vector is directed tangent to the circle as well.. An object moving in a circle is accelerating. Accelerating objects are objects which are changing their velocity - either the speed (i.e., magnitude of the velocity vector) or the direction. An object undergoing uniform circular motion is moving with a constant speed. theless, it is accelerating due to its change in direction. The direction of the acceleration is inwards. The animation at the right depicts this by means of a vector arrow. The final motion characteristic for an object undergoing uniform circular motion is the net force. The net force acting upon such an object is directed towards the center of the circle. The net force is said to be an inward orcentripetalforce. Without such an inward force, an object would continue in a straight line, never deviating from its direction. Yet, with the inward net force directed perpendicular to the velocity vector, the object is always changing its direction and undergoing an inward acceleration. |
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