InterviewSolution
Saved Bookmarks
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.
| 1101. |
A body is desplaced from origin to `(2m, 4m)` under the following two forces: (a) `F=(2^hati + 6^hatj)N`, a constant force (b) `F(2x^hati + 3y^(2)hatj)N` Find word done by the given forces in both cases. |
|
Answer» Correct Answer - B (a) `F=(2hati + 6hatj)N` `dr=(dxhati + dyhatj)` `:. fdr= 2 dx + 6 dy` `W =int_((0.0))^((2m.4m))F.dr = int_((0.0))^((2m.4m))(2dx + 6dy)` `=[2x + 6y]_(0,0)^(2m,4m) =(2xx2 + 6xx4)` `=28J` |
|
| 1102. |
The tangenital acceleration of a particle in a circular motion of radius 2 m is `a_(t)=alphat m//s^(2)` (where `alpha` is a constant) initially the particle as rest. Total acceleration of the particle makes `45^(@)` with the radial acceleration after 2 sec The value of constant `alpha` is :A. `1/2 m//s^(3)`B. `1 m//s^(3)`C. `2 m//s^(3)`D. Data are insufficient |
|
Answer» Correct Answer - B |
|
| 1103. |
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration `a_(c)` is varying with time t as `a_(c) = k^(2)rt^(2)`, where k is a constant. The power delivered to the particle by the forces acting on it is :A. `2 pimk^(2)r^(2)t`B. `mk^(2)r^(2)t`C. `1/3 mk^(4)r^(2)t^(5)`D. 0 |
|
Answer» Correct Answer - B |
|
| 1104. |
A particle of mass m impves along a horzontal circle of radius R such that normal acceleration of particle varies with time as `a_(n)=kt^(2).` where k is a constant. Tangential force on particle at t s isA. `2msqrtkR`B. `msqrtkR`C. `m/2sqrt(kR)`D. `msqrt((kR)/(2))` |
| Answer» Correct Answer - B | |
| 1105. |
A particle of mass m impves along a horzontal circle of radius R such that normal acceleration of particle varies with time as `a_(n)=kt^(2).` where k is a constant. Power developed by total at time t isA. `(mkRT)/(3)`B. `2mkRT`C. `(mkRT)/(2)`D. mkRT |
| Answer» Correct Answer - D | |
| 1106. |
A particle of mass m impves along a horzontal circle of radius R such that normal acceleration of particle varies with time as `a_(n)=kt^(2).` where k is a constant. Total force on particle at time t s isA. `msqrt(k(R^(2)+kt^(4)))`B. `msqrt(k(R+kt^(4)))`C. `2msqrt(k(R+kt^(4)))`D. `m/2sqrt(k(R+kt^(4)))` |
| Answer» Correct Answer - B | |
| 1107. |
A force `F=-k(^hati + x^hatj)` (where k is a positive constant) acts on a particle moving in the `x-y` 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 particle is (a) `-2ka^(2)` , (b) `2ka^(2) , (c)`-ka^(2)` , (d) `ka^(2)` |
|
Answer» Correct Answer - A::B::C `dW=F.dr`, where `dr=dxhat(i) + dyhat(j) + dzhat(k)` and `F=k(yhati + xhatj)` `:. dW=-k(ydx + xdy)=-kd(xy)` `:. W=-int_(0,0)^(a,a)dW =-kint_(0,0)^(a,a)d(xy)` `=-k[xy]_(0,0)^(a,a)`, `W=-ka^(a)` `:.` The correct option is `(c)`. |
|
| 1108. |
A point moves along a circle with a speed `v=kt` , where `k=0.5m//s^(2)` Find the total acceleration of the point the moment when it has covered the `n^(th)` fraction of the circle after the begining of motion, where `n=(1)/(10)` . |
|
Answer» Correct Answer - B |
|
| 1109. |
A force `F=-k(y hati + x hatj)` (where k is a positive constant) acts on a particle moving in the `x-y` 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 particle is (a) `-2ka^(2)` , (b) `2ka^(2) , (c)`-ka^(2)` , (d) `ka^(2)`A. `-2ka^2`B. `2ka^2`C. `-ka^2`D. `ka^2` |
|
Answer» Correct Answer - C `dW=vecF.dvecr=-k(yhati+xhatj)xx(dxhati+dyhatj)` `=-k[ydx+xdy]=-kd(xy)` `W=-kint_(0.0)^(a,a)d(xy)=-k|xy|_(0,0)^(a,a)` |
|
| 1110. |
A particle moves along the X-axis from `x=0 to x=5m` under the influence of a force given by `F=(7-2x+3x^(2)).` Find the work done in the process. |
|
Answer» Here, `F=7-2x+3x^(2).` From `x=0 to x=5m, ` work done by the force. `W=int_(0)^(5)Fdx=int_(0)^(5)(7-2x+3x^(2))dx` `=[7x-(2x^(2))/(2)+(3x^(3))/(3)]_(0)^(5)` `=7(5-0)-(5^(2)-0^(2))+(5^(3)-0)` `=135J` |
|
| 1111. |
A particle moves along the X-axis from `x=0 to x=5m` under the influence of a force given by `F=(7-2x+3x^(2)).` Find the work done in the process.A. 70B. 270C. 35D. 135 |
|
Answer» Correct Answer - D `W=int_(x1)^(x2)Fdx=int_(0)^(5)(7-2x+3x^2)dx` `=|7x-x^2+x^3|_0^5` `=7(5)-(5)^2(5)^3=35-25+125` `=135J` |
|
| 1112. |
Statement-1 : Energy released when a mass of one microgram disappears in a process is `9xx10^(7)J`. Statement-2 : It follows from `E=(1)/(2)m upsilon^(2)`A. Statement - 1 is ture , statement -2 is true, and statement-2 is correct explanation of statement -1.B. Statement-1 if ture, statement-2 is true, but statement -2 is not a correct explanation of statement-1.C. Statement-1 is true, but statement -2 is false.D. Statement-1 is false, but statement-2 is true. |
|
Answer» Correct Answer - C From `E=mc^(2)`, when `m=1` microgram`=10^(6)g=10^(-9)kg` `E=10^(-9)(3xx10^(8))^(2)=9xx10^(7)J` The statement`-1` is true, but the statement `-2` is certainly false. Choice (c) is correct. |
|
| 1113. |
A particle is acted by `x` force `F = Kx` where `K` is `a( + Ve)` constant its potential mwrgy at `x = 0` is zero . Which curve correctly represent the variation of putential energy of the block with repect to `x`A. (a) B. (b) C. (c) D. (d) |
|
Answer» Correct Answer - B We know that `DeltaU=-W` for conservative forces `:. DeltaU=-underset0oversetx intFdx=-underset0oversetx int Kxdx` `implies U_((x))-U_((0))=-(kx^2)/(2)` Given `U_((0))=0`. Therefore, `U_((x))=-(kx^2)/(2)` |
|
| 1114. |
The potential energy of a particle of mass `m` free to move along the x-axis is given by `U=(1//2)kx^2` for `xlt0` and `U=0` for `xge0` (x denotes the x-coordinate of the particle and k is a positive constant). If the total mechanical energy of the particle is E, then its speed at `x=-sqrt(2E//k)` isA. (a) ZeroB. (b) `sqrt((2E)/(m))`C. (c) `sqrt(E/m)`D. (d) `sqrt((3E)/(2m))` |
|
Answer» Correct Answer - A From the conservation of energy `KE+PE=E` or `KE=E-1/2kx^2` `KE` at `x=-sqrt((2E)/(k))` is `E-1/2k((2E)/(k))=0` The speed of particle at `x=-sqrt((2E)/(k))` is zero. |
|
| 1115. |
You are reshelving books in a library. You lift a book from the floor to the top shelf. The kinetic energy of the book on the floor was zero and the kinetic energy of the book on the top shelf is zero, so no change occurs in the kinetic energy even though you did some work in lifting the books. Is the work-kinetic energy theorem violated? |
| Answer» There is no violation. Choose the book as the system. You did work and the earth did work on the book. The average force you exerted just counterbalanced the weight of the book. The total work on the book is zero, and is equal to its overall change in kinetic energy, which is also zero. | |
| 1116. |
A well `20m` deep and `3m` is diameter contains water to a dept of `14m`. How long will a `5 hp` engine take to empty it ? |
|
Answer» Here, depth of well `=20m, `radius `r=(3)/(2)m` depth of water `=14m,t=?` `P=5hp=5xx746W` Area of cross section of well, `A=pir^(2)=(220/(7))(3//2)^(2)=(99)/(14)m^(2)` Volume of water in well, `V=Axxd=(99)/(14)xx14=99m^(3)` Mass of water in well, `m=Vxxrho=99xx10^(3)kg` When the well is emptied, depth of water changes from `(20-14)m` in the beginning to `920-14)m` at the end. `:.` Average height raised, `h=(6+20)/(2)=13m` Work done, `W=mgh==99xx10^(3)xx9.8xx13` As `P=(W)/(t)`, `t=(W)/(P)=(99xx10^(3)xx9.8xx13)/(5xx746)` `3381.6se cond` |
|
| 1117. |
A car of mass `M` accelerates starting from rest. Velocity of the car is given by `v=((2Pt)/(M))^((1)/(2))` where `P` is the constant power supplied by the engine. The position of car as a function of time is given as |
|
Answer» Correct Answer - A |
|
| 1118. |
A pump can take out 7200 kg of water per hour from a well 100 m. deep. The power of pump, assuming its efficiency as `50%` will beA. 1 kWB. 2 kWC. 3 kWD. 4 kW |
|
Answer» Correct Answer - D Output Power`=(7200xx10xx100)/(3600)=2000W` `efficiency=(output power)/(i ntput power)` `0.5=(200)/(i nputpower)` `i nput power=4000W=4kW` |
|
| 1119. |
The power of pump, which can pump 200 kg of water to height of 200 m in 10 s `(g=10(m)//(s^2)`)A. 40 kWB. 80 kWC. 400 kWD. 960 kW |
|
Answer» Correct Answer - A `overlineP=(mgh)/(t)=((200)(10)(200))/(3600)=40000W` `=40kW` |
|
| 1120. |
The power of pump, which can pump 200 kg of water to a height of 50 m in 10 sec, will beA. `4xx10^(3) W`B. `10xx10^(3)W`C. `20xx10^(3)W`D. None of these |
|
Answer» Correct Answer - A (b) We know that `"power", P=(W)/(t)=(mgh)/(t)=(200xx10xx50)/(10)=10xx10^(3)W` |
|
| 1121. |
An automobile of mass m accelerates from rest. If the engine supplies a constant power P, the velocity at time t is given by -A. `(2Pt)/(M)`B. `sqrt((2Pt)/(m))`C. `(Pt)/(2M)`D. `sqrt((Pt)/(2 M))` |
|
Answer» Correct Answer - B (b) We have, P=Fv=constant (given) `rArr M((dv)/(dt))v=P" or ""vdv"=(Pdt)/(M)` Integrating both side, we get `(v^(2))/(2)=(P)/(M)t" or "v=sqrt((2Pt)/(M))` |
|
| 1122. |
A particle which is constant to move along the `x- axis` , is subjected to a force in the same direction which varies with the distance `x` of the particle from the origin as `F(x) = -Kx + ax^(3)` . Hero `K` and `a` are positive constant . For `x ge 0`, the fanctional from of the patential every `U(x) of the particle isA. B. C. D. |
|
Answer» Correct Answer - D `F = (-dU)/(dx) rArr dU = - F dx` `rArr U = - int_(0)^(x) (-kx +ax^(3)) dx = (kx^(2))/(2) - (ax^(2))/(4)` `:. `We get `U =0` at `x= 0` and `x= sqrt(2k//a)` So `F = 0` at `x= 0` i.e. , slope of `U` - x graph is at `x= 0` |
|
| 1123. |
A block of mas 2.0 kg is pulled up on a smooth incline of angle `30^0` with the horizontal. If the block moves wth an acceleration of `1.0 m/s^2`, find the power delivered by the pulling force at a time 4.0 s after the motion starts. What is the average power delivered during the 4.0 s after the motion starts? |
|
Answer» Here, the given plane is smooth. Therefore, there is no force of friction. Now, `m=2kg,theta=30^(@), a=1m//s^(2),P=?` Force F applied up the plane is such that from figure., `F-mg sin theta =ma` `F=m(g sintheta +a)` `=2(9.8 sin 30^(@) +1)` `F=11.8N` From `v=u+at` `v=0+1xx4=4m//s` `:.` Power delivered by the force at a time 4 second after the motion starts is `P=Fxxv=11.8xx4=47.2 wat t` Now, displacement in 4 second is `s=ut+(1)/(2)at^(2)=0+(1)/(2)xx1xx4^(2)=8m` work done `= Fxxs=11.8xx8=94.4J` Average power delivered during 4 second after the motion starts `P=(W)/(t)=(94.4)/(4)=23.6"watt"` |
|
| 1124. |
An ideal massless spring is compressed by 1 m by a force of 100 N. The same spring is placed at the bottom of a frictionless inclined plane `(theta = 30^(@))`. A 10-kg mass is released from the top of the incline and is brought to rest momentarily after compressing the spring by 2m. Through what distance does the mass slide before coming to rest ? [Hint : Apply work-energy theorem]A. `sqrt(20)m//s`B. `sqrt(30)m//s`C. `sqrt(10)m//s`D. `sqrt(40)m//s` |
|
Answer» Correct Answer - A |
|
| 1125. |
An ideal massless spring `S` can compressed `1.0` m in equilibrium by a force of `100 N`. This same spring is placed at the bottom of a friction less inclined plane which makes an angle `theta =30^@` with the horizontal. A (10 kg) mass (m) is released from rest at the top of the incline and and is brought to rest momentarily after compressing the spring by `2 m. if `g =10 ms^(-2)`, what is the speed of just before it touches the spring? . |
|
Answer» Correct Answer - [(a) 4m, (b) `sqrt(20)m//s`] |
|
| 1126. |
A brick is placed on a vertical compressed massless spring of force constant `350N//m` attached to the ground. When the spring is released the brick is propelled upward. If the brick mass is 1.8 kg and will reach a maximum height of 3.6 m above its initial position on the compressed spring, what distance must the spring be compressed initially ? Take `g=10m//s^(2)` |
|
Answer» Correct Answer - [0.608 m] |
|
| 1127. |
The spring is compressed by a distance a and released. The block again comes to rest when the spring is elongated by a distance `b`. During this A. Work done by the spring on the block `=1/2k(a+b)^(2)`B. Work done by the spring on the block `=1/2 k(a^(2)-b^(2))`C. Coefficient of friction `=(k(a-b))/(2mg)`D. Coefficient of friction `=(k(a+b))/(2mg)` |
|
Answer» Correct Answer - B::C |
|
| 1128. |
A bullet, moving with a speed of `150(m)//(s)`, strikes a wooden plank. After passing through the plank, its speed becomes `125(m)//(s)`. Another bullet of the same mass and size strikes the plank with a speed of `90(m)//(s)`. Its speed after passing through the plank would beA. `25(m)//(s)`B. `35(m)//(s)`C. `50(m)//(s)`D. `70(m)//(s)` |
|
Answer» Correct Answer - B `(1)/(2)m(150)^2=(1)/(2)m(125)^2+Rd` .(i) `(1)/(2)m(90)^2=(1)/(2)mv^2+Rd` .(ii) `(150)^2-(125)^2=(90)^2-v^2` `v=35(m)//(s)` |
|
| 1129. |
A block is projected with a speed `v_(0)` strikes the point pf projection after describing the path as shown in figure by dotted line. If friction exists for the path of length d and the vertical circular path is smooth, assuming coefficient of frictioin to be `mu` (A) Find `v_(0)`? (B) What is the minimum value of `v_(0)` ? |
| Answer» `(A)sqrt((d^(2)g)/(4R)+2g(mud+2R)),(B)sqrt((4R+2mud))g` | |
| 1130. |
A block of mass `m` is released from rest onto a spring. A having stiffness `k_A=mg//2h` as shown in figure. If the block compresses spring B through a distance h, find the: a. stiffness of the spring B b. equilibrium position of the block c. maximum velocity of the block d. maximum acceleration of the block |
| Answer» `(A)(4mg)/(h),(B)h/g` below top of spring B, `(C) y=4/3sqrt(2gh)(D)4guarr` | |
| 1131. |
A `16kg` block moving on a frictionless horizontal surface with a velocity of `4m//s` compresses an ideal spring and comes to rest. If the force constant of the spring be `100N//m`, then how much is the spring commpressed ? |
|
Answer» `K.E. ` of block `=` work done in compressing the spring `(1)/(2)m upsilon^(2)=(1)/(2)k x^(2)` `x=sqrt((m upsilon^(2))/(k))=upsilonsqrt((m)/(k))=4sqrt((16)/(100))=1.6m` |
|
| 1132. |
Figure shows two smooth inclined planes with different inclinations. Two blocks of same mass are allowed to slide down the two planes from the top A. Which block will arrive on the ground with greater velocity? |
|
Answer» As is clear from figure, both the blocks slide down different slopes from the same height h. As there is no friction and gravitational froces are conservative. `:. ` K.E. on the ground `=` P.E. at the top `(1)/(2)mv^(2)=mgh` `v=sqrt(2gh)` `:.` Both the blocks will reach the ground with the same velocity. |
|
| 1133. |
A block Q of mass 2m is placed on a horizontal frictionless plane. A second block of mass m is placed on it and is connected to a spring of spring constant K, the two block are pulled by distance A. Block Q oscillates without slipping. The work done by the friction force on block Q when the spring regains its natural length is: A. `1/3 KA^(2)`B. `2/3 KA^(2)`C. `1/2 KA^(2)`D. `1/4 KA^(2)` |
|
Answer» Correct Answer - A |
|
| 1134. |
Two blocks A and B of mass m and `2m` respectively are connected by a light spring of force constant k. They are placed on a smooth horizontal surface. Spring is stretched by a length x and then released. Find the relative velocity of the blocks when the spring comes to its natural length A. `(sqrt((3k)/(2m)))x`B. `(sqrt((2k)/(3m)))x`C. `sqrt((2kx)/m)`D. `sqrt((3km)/(2x)` |
|
Answer» Correct Answer - A |
|
| 1135. |
Work is said to be done by a force acting on a body, provided the body is displaced actually in ay direction except in a direction perpendicular to the direction of the force. Mathmatically, `W=vec(F). vec(c)=Fs cos theta`, Whereas eenrgy is capacity of a body to do the work, power is the rate at which the body can do the work. `P=(W)/(t)=vec(F).(vec(s))/(t)=vec(F). vec(upsilon)` Both, work and energy are measured in joule and power is measured in watt. With the help of the comprehension given above, choose the most appropriate alternative for each of the following question `:` In the above question, work done by gravity isA. `400 J`B. `-400 J`C. zeroD. `-25 J` |
|
Answer» Correct Answer - C As motion is along horizontal and gravity is along the vertical, therefore, work done by gravity is `W=F s cos 90^(@)=Zero` |
|
| 1136. |
A force `F=(5+2x)` acts on a particle in x direction where F is in newtown and x is meter. Find the work done by this force during displacement from `x=0` to `x=1m.` |
|
Answer» `F=(5+2x)` `dW=vecF.vec(dx)=(5+2x)dx` Total work done,` W=underset(x=0)overset(x=1)int(5+2x)dx` `F=(5+2x)` `dW=vecF.vec(dx)=(5+2x)dx` `W=[5x+2.(x^(2))/(2)]_(0)^(1)` `=5(1-0)+(1^(2)-0)=6J` |
|
| 1137. |
Work is said to be done by a force acting on a body, provided the body is displaced actually in ay direction except in a direction perpendicular to the direction of the force. Mathmatically, `W=vec(F). vec(c)=Fs cos theta`, Whereas eenrgy is capacity of a body to do the work, power is the rate at which the body can do the work. `P=(W)/(t)=vec(F).(vec(s))/(t)=vec(F). vec(upsilon)` Both, work and energy are measured in joule and power is measured in watt. With the help of the comprehension given above, choose the most appropriate alternative for each of the following question `:` `In the above question, work done by the resisting force isA. `400 J`B. `-400 J`C. zeroD. `-25 J` |
|
Answer» Correct Answer - B Resisting force opposes the applied force. Box moves at `180 ^(@)` to the resisting force. `:. W= F cos theta= 100 xx 4 cos 180 ^(@)=-400J` |
|
| 1138. |
A force `F = Ay^(2)+By+C` acts on a body at rest in the `Y`-direction. The kinetic energy of the body during a displacement `y = -a` to `y = a` isA. `(2Aa^(3))/(3)`B. `(2Aa^(3))/(3)+2Ca`C. `(2Aa^(3))/(3)+(Ba^(2))/(2)+Ca`D. None of these |
|
Answer» Correct Answer - B (b) `W=int_(-a)^(+a)Fdy=[(Ay^(3))/(3)+(By^(2))/(2)+Cy]_(-a)^(+a)=(2Aa^(3))/(3)+2Ca` |
|
| 1139. |
The power of a water pump is 2 kW. If `g = 10 m//s^2,` the amount of water it can raise in 1 min to a height of 10 m is :A. 2000 litreB. 1000 litreC. 100 litreD. 1200 litre |
|
Answer» Correct Answer - D |
|
| 1140. |
An engine develops 10 kW of power. How much time will it take to lift a mass of 200 kg to a height of 40 m (`g=10(m)/(sec^2)`)`?A. 4 secB. 5 secC. 8 secD. 10 sec |
|
Answer» Correct Answer - C |
|
| 1141. |
The average power required to lift a 100 kg mass through a height of 50 metres in approximately 50 seconds would beA. 50J/sB. 5000 J/sC. 100J/sD. 980 J/s |
|
Answer» Correct Answer - D |
|
| 1142. |
Heavy water is used as moderator in a nuclear reactor. The function of the moderator isA. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If asserti on is true but reason is falseD. If the assertion and reason both are false |
|
Answer» Correct Answer - C |
|
| 1143. |
you lift a heavy book from the floor of the room and keep it in the book - shelf having a height `2m` in this process you take `5seconds` The work done you will depend uponA. Mass of the book and time takenB. Weight of the book and height of the book-shelfC. Height of the book-shelf and time takenD. Mass of the book, height of the book-shelf and time taken |
|
Answer» Correct Answer - B Option b is correct Work depends on Force(F) and displacement(s). Here F=wieght of the book S= height of the book shelf. |
|
| 1144. |
These questions consists of two statements each printed as Assertion and Reason. While answering these question you are required to choose any one of the following five reponses (a) If both Assertion and Reason arecorrect and Reason is the correct explanation of Asserrtion. (b) If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion. (c ) If Assertion is true but Reason is false. (d) If Assertion is false but Reason is true. Assertion Spring force is a conservative force. Reason Potential energy is defined only for conservative forces. |
|
Answer» Correct Answer - A (d) Spring force is conservative in nature. Potential energy is only associated with conservative force. |
|
| 1145. |
These questions consists of two statements each printed as Assertion and Reason. While answering these question you are required to choose any one of the following five reponses (a) If both Assertion and Reason arecorrect and Reason is the correct explanation of Asserrtion. (b) If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion. (c ) If Assertion is true but Reason is false. (d) If Assertion is false but Reason is true. Assertion A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of `30^(@)` with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in the mechanical energy in the second situation is smaller than that in the first situation. Reason The coefficient of friction between the block and the surface decreases with the increase in the anglle of inclination. |
|
Answer» Correct Answer - C (c ) Assertion Decrease in mechanical energy in case 1 will be `DeltaU_(1)=(1)/(2)mv^(2)` But decrease in mechanical energy in case 2 will be `DeltaU_(2)=(1)/(2)mv^(2)-mgh` `:. " " DeltaU_(2) lt DeltaU_(1)` or assertion is correct. Coefficient of friction is the mutual property of two surfaces. It does not depend on angle of inclination. |
|
| 1146. |
These questions consists of two statements each printed as Assertion and Reason. While answering these question you are required to choose any one of the following five reponses (a) If both Assertion and Reason arecorrect and Reason is the correct explanation of Asserrtion. (b) If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion. (c ) If Assertion is true but Reason is false. (d) If Assertion is false but Reason is true. Assertion. Velocity of a block changes from `(2hati+3hatj) ms^(-1)` to `(-4hati-6hatj) ms^(-1)`. Then, work done by all the forces during this interval of time is positive. Reason Speed of block is increasing. |
|
Answer» Correct Answer - A (a) According to work-energy theorem, work done by all the forces =change in kinetic energy. |
|
| 1147. |
1 a.m.u is equivalent toA. `1.6xx10^(-12)`jouleB. `1.6xx10^(-19)` jouleC. `1.5xx10^(-10)` jouleD. `1.5xx10^(-19)` joule |
|
Answer» Correct Answer - C |
|
| 1148. |
A block of mass m initially at rest is dropped from a height h on to a spring of force constant k . the maximum compression in the spring is x thenA. `mgh=(1)/(2)kx^(2)`B. `mg(h+x)=(1)/(2)kx^(2)`C. `mgh=(1)/(2)k(x+h)^(2)`D. `mg(h+x)=(1)/(2)k(x+h)^(2)` |
|
Answer» Correct Answer - B |
|
| 1149. |
Concerete block of mass `m_(A) and m_(B)`. The gravitutional potential energy of each system is zero at the equilibrium position of the springs. Which ststement is true for the total machanical energy of the springs. Which system when the block , are balanced on the springs?A. `E_(A) = E_(B)`B. `E_(A) = 2 E_(B)`C. `E_(A) = 4 E_(B)`D. `E_(A) = -2 E_(B)` |
|
Answer» Correct Answer - C `Kx_(A) = m_(A) g` `Kx_(B) = m_(B) g implies (x_(A))/(x_(B)) = (m_(A))/(m_(B)) = 2 implies x_(A) = 2 x_(B)` `PE_(A) = (1)/(2) K x_(A)^(2) PE_(B) = K x_(B)^(2) implies (PE_(A))/(PE_(B)) = ((x_(A))/(x_(B)))^(2) = 4` So for `A`, potential energy of spring is four times that for `B`. So `E_(A) = 4E_(B)` |
|
| 1150. |
Two springs have their force constant as `k_(1)` and `k_(2) (k_(1) gt k_(2))`. When they are streched by the same force.A. net work is done in case of both the same springsB. Equal work is done in case of both the springsC. More work is done in case of both the second springsD. More work is done in case of both the first springs |
|
Answer» Correct Answer - C `W=(F^(2))/(2k)` If both springs are stretched by same force then `W=prop(1)/(2)` As `k_(1)gt k_(2)` therefore `W_(1) lt W_(2)` i.e, more work is done in case of second spring. |
|