HomePhysics Class 11WORK ENERGY AND POWER MCQ TEST 1 WORK ENERGY AND POWER MCQ TEST 1 Work Engergy and Power 11 CBSE - MCQ test 1 Question 1. --------------product of force and dispalcement is called workdone. Cross Dot Explanation Answer : Dot product Question 2. Workdone is a ...............physical quantity. Vector Scalar Explanation Answer: Scalar Work done is a dot product of vectors force and the displacement. Since, the dot product is a scalar quantity. Question 3. What is dimensional formula of workdone? [M²L²T¯²] [ML²T¯²] [MLT¯²] None Explanation Answer: [ML²T¯²] The dimensional formula of work done is [ML²T¯²]. Question 4. The magnitude of work done by a force: depends on frame of reference does not depend on frame of reference cannot be calculated in non-inertial frames. both (a) and (b) Explanation Answer : depends on frame of reference The work done by a force is dependent on the frame of reference. Question 5. Work done by a force on an object is zero when : displacement is zero force is zero force and displacement are mutually perpendicular All of these Explanation Answer:All of these W=Fdcosθ When d=0,W=0 When F=0,W=0 When θ=90°,W=0 Question:6 When the force acting on body retards the motion of body, the work done will be: negative positive zero None Explanation Answer: Negative Force is acting opsite of direction of velocity or displacement The angle between force and displacement is 180°. Question:7 Work is always done on a body when: a force acts on it it moves through a certain distance it experiences an increase in energy through a mechanical influence. None of these Click Here For Solution Answer: it experiences an increase in energy through a mechanical influence. According to work energy theorem work done on body =change in kinetic energy of the body Question:8 A force of 50 N acts on a body and moves it through a distance of 4 m on a horizontal surface. What would be work done in Joules if the direction of force is at an angle of 60∘ to the horizontal surface. 200J 150J 100 J None Explanationn Answer:100j Given: =>Force (F) = 50 N =>Distance (d) = 4 m =>Angle θ= 60∘ =>Therefore, W=50×4×1/2 (since cos 60∘ = ½) =>W=100 J Question:9 Which of following is correct ? Workdone= (force)*(displacement) Workdone= (force)*(component of displacement in direction of force) Workdone= (displacement)*(component of force in direction of displacement) All of these Expanation Answer:All of these W=Fdcosθ When θ=90°,W=Fd [Workdone= (force)*(displacement)] W=F(dcosθ)= (force)*(component of displacement in direction of force) W=d(Fcosθ)= (displacement)*(component of force in direction of displacement) Question:10 If the units of length and force are increased three times, then unit of energy will becomes 6 times. becomes 8 times. becomes 9 times. None Explanation Answer:becomes 9 times. W=Fdcosθ When F1=3F,d=3d W1=3F*3dcosθ W1=9Fdcosθ W1=9W Question:11 A forceF =3i +cj +2k N acting on a particle causes a displacement S =−4i +2j −3k m. lf the work done is 6 Joule, the value of c is: 0 1 12 None Explanation Answer:12 Question:12 A body of mass 6kg is under a force which causes displacement in it given by S= t²/4 metres where t is time. The work done by the force in 2 seconds is: 12J 9J 3J None of these Explanation Answer:3J Given,S= t²/4 v=ds/dt=2t/4=t/2 v1 at t=0=>0 v2 at t=2=>1 Workdone= change in KE =1/2*6*1-0=3J Question:13 A particle moves from position r1=3i +2j −6k to position r2=14i +13j +9k under the force 4i +j +3k . Find the work done. 10J 100J 150J None Explanation Answer:100J Question:14 The work done against gravity in taking 10kg mass at 1m height in 1s will be (in J) 196J 49J 98J None of these Explanation Answer:98J Work done against gravity =mgh=10×9.8×1=98J Question:15 A force acting on a particle varies with the displacement x as F=ax−bx². Where a=1 N/m and b=1 N/m². The work done by this force for the first one meter (F is in newtons, x is in meters) is : (3/6)J (2/6)J (1/6)J None Explanation Answer:(1/6)J Question:16 A force F=(3xi +4j ) Newton (where x is in metres) acts on a particle which moves from a position (2 m, 3 m) to (3 m, 0 m). Then the work done is : -4.5J -7.5J 16J None of these Explanation Answer:-4.5J Question:17 The position of a particle varies with time as x = (t-2)², where x is in meters and t is in seconds. Calculate the work done during t =0 to t = 4s , if mass of the particle is 100g. 0.7J 0.5J 0 None of these Explantion Answer:0 Question:18 The minimum work done in slowly pulling up a block of wood weighing 2kN for a length of 10m on a smooth plane inclined at an angle of 15° with the horizontal by a force parallel to the incline is : (sin15°=0.26,cos15°=0.96) 4.36KJ 5.36KJ 3.36KJ 5.17KJ Click Here For Solution Weight of the block (W=mg)=2 kN = 2000 N Length of the incline (d) = 10 m Angle of the incline (θ) = 15° Height raised =10sin15° Work done = Change in potential energy of block ⇒W=mgh ⇒W= 2000*10sin15°= 5.17 KJ Question:19 Unit of energy is: kwh joule electron volt All of the above Click Here For Solution Answer:All of the above Question:20 A ball moves in a frictionless inclined table without slipping. The work done by the table surface on the ball is Negative Zero Positive None of these Explanation Answer:Zero Workdone =0 because there no force acting on ball by the table. 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