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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Match the following: `("Gravitational potential energy" (U=0, "at the reference level"))/g` (in `J-s^(2)//m`) of the body (shown in column-1) with respect to the reference level (shown in column-2) where `g` is gravitation acceleration.A. (I)(i)(S)B. (II)(ii)`(R)`C. (III)(iii)(P)D. (IV)(iv)(Q) |
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Answer» Correct Answer - D `U/g=Mh_(CM)=(2)(3/2)=2m` |
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| 2. |
A block of mass `m=15g` is suspended in an elevator with the help of three identical light elastic cords (spring constant `k=100N//m` each) attached vertically. One of them cord 1 is tied to the ceilling of the elevator and the other two cords 2 and 3 are tied to the elevator floor as shown in the figure. When the elevator is stationary the tension force is each of the lower cords is `T=7.5N`. Take `g=10m//s^(2)`. Now the elevator starts moving with given four accelerations shown in column I. Column II given the displacement of the block with respect to elevator when its is accelerating. Column III gives tension in the cords when elevator is accelerating. Tension in cord -2A. (I)(iv)(P)B. (II)(i)(Q)C. (III)(iii)`(R)`D. (IV)(ii)(S) |
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Answer» Correct Answer - A For elevator acceleration `1m//s^(2)`, tension in cord 2 and cord 3 decreases. So `165+DeltaT-2xx(7.5-DeltaT)-150=(15)(1)` `implies DeltaT=5N` `implies` Tension in cord 2, `T_(2)=2.5N` |
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| 3. |
A block of mass `m=15g` is suspended in an elevator with the help of three identical light elastic cords (spring constant `k=100N//m` each) attached vertically. One of them cord 1 is tied to the ceilling of the elevator and the other two cords 2 and 3 are tied to the elevator floor as shown in the figure. When the elevator is stationary the tension force is each of the lower cords is `T=7.5N`. Take `g=10m//s^(2)`. Now the elevator starts moving with given four accelerations shown in column I. Column II given the displacement of the block with respect to elevator when its is accelerating. Column III gives tension in the cords when elevator is accelerating. Tension in cord -1A. (I)(i)(S)B. (II)(i)`(R)`C. (III)(iii)(P)D. (IV)(iii)(Q) |
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Answer» Correct Answer - B::C For elevator acceleration `1.5m//s^(2)`, tension is cord 2 and cord 3 is zero So, `T_(1)-150=(15)(1.5)` `implies` Tension in cord `1, T_(1)=172.5N` |
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| 4. |
A block of mass `m=15g` is suspended in an elevator with the help of three identical light elastic cords (spring constant `k=100N//m` each) attached vertically. One of them cord 1 is tied to the ceilling of the elevator and the other two cords 2 and 3 are tied to the elevator floor as shown in the figure. When the elevator is stationary the tension force is each of the lower cords is `T=7.5N`. Take `g=10m//s^(2)`. Now the elevator starts moving with given four accelerations shown in column I. Column II given the displacement of the block with respect to elevator when its is accelerating. Column III gives tension in the cords when elevator is accelerating. Tension in cord -3A. (I)(i)(P)B. (II)(ii)(P)C. (III)(iii)(S)D. (IV)(iii)`(R)` |
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Answer» Correct Answer - C For elevator acceleration `2m//s^(2)` tension in cord 2 and cord 3 is zero. So, `T_(1)-150=(15)_(2)` `implies` Tension in cord `1,T_(1)=180N` So displacement of block `=15cm` downward |
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| 5. |
what are the benefits of inflation? |
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Answer» Inflation means the value of money will fall and purchase relatively fewer goods than previously. Inflation allows borrowers to pay lenders back with money worth less than when it was originally borrowed, which benefits borrowers. When inflation causes higher prices, the demand for credit increases, raising interest rates, which benefits lenders. |
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| 6. |
supply function is Q= -100+10P . Where price = 15 . find elasticity by point method . |
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Answer» Given supply function is given as: Q = -100 + 10P => P = (Q + 100)/10 = Q/10 + 100/10 = Q/10 + 10. Slope of supply = 1/10. When price (P) = 15, Q = -100 + 10(15) = -100 + 150 = 50. Elasticity of supply = (1/Slope of supply) x P/Q = {1/(1/10)} x 15/50 = 10 x 15/50 = 150/50 = 3. |
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| 7. |
A power output from a certain experimental car design to be shaped like a cube is proportional to the mass `m` of the car. The force of air friction on the car is proportional to `Av^(2)`, where `v` is the speed of the car and `A` is the cross-sectional area. On a level surface the car has a maximum speed `v_("max")`. Assume that all versions of this design have the same density. The `v_("max")` is propotional to `m^(1//c)`. Find `C` |
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Answer» At `v_("max")` `Av_("max")^(2)=F=(km)/(v_("max"))` (where `k` is a propotionality constant) `m^(2//3)v_("max")^(2) prop m/(v_("max"))` `:.v_("max")prop m^(1//9)` |
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| 8. |
A particle is projected with speed `10 m//s` at angle `60^(@)` with the horizontal. Then the time after which its speed becomes half of initial.A. `(1)/(2)sec`B. `1sec`C. `sqrt(3//2)sec`D. `sqrt(3//2)sec` |
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Answer» Correct Answer - D `u cos60^(@)=5,V_(y)=usin 60^(@)-10t` `V_(2)=(u sin 60^(@)-10t)_(2)+(u cos 60^(@))` `(u^(2))/(4)=(u(sqrt(3))/(2)-10t)^(2)+(u^(2))/(4)` `rArr 10t=(10 sqrt(3))/(2) rArr t=(sqrt(3))/(2)`. |
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| 9. |
A sphere of radius R and density `rho_1` is dropped in a liquid of density `sigma`. Its terminal velocity is `v_1`. If another sphere of radius `R` and density `rho_2` is dropped in the same liquid, its terminal velocity will be:A. `v[p_(2)+sigma)//p_(1)+sigma)]`B. `v[p_(1)+sigma)//p_(2)+sigma)]`C. `v[p_(2)-sigma)//p_(1)-sigma)]`D. `v[p_(1)-sigma)//p_(2)-sigma)]` |
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Answer» Correct Answer - C The terminal velocity of the sphere `v =(2)/(9)R^(2) ((rho^(2) - sigma)g)/(eta)" ".....(i)` Here ,R = radius of sphere `rho_(1)` = density of sphere and `sigma` = density of liquid . Now if density of metal sphere is changed to `rho_(2)` then terminal velocity `v_(2) (2)/(9)(R^(2) (rho_(2) - sigma)g)/((rho_(1) - sigma)) or (v_(2)(rho_(2) - sigma))/((rho_(1) - sigma))` Hence , the terminal velocity is `(v(rho_(2)-sigma))/(rho_(1)-sigma)` |
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| 10. |
Graphical soluton of a two body head on collision A block `A` of mass `m` moving with a uniform velocity `v_(0)` strikes another identical block `B` kept at rest on a horizontal smooth surface as shown in the figure (i). We can conserve linear momentum. So `mv_(0)=mv_(A)mv_(B)` (`v_(A)` and `v_(B)` are the velocities of the blocks after collision) `:. v_(0)=v_(A)+v_(B)`.........(i) If the collision is perfectly elastic `1/2 mv_(0)^(2)=1/2 mv_(A)^(2)+1/2 mv_(B)^(2)` `impliesv_(0)^(2)=v_(A)^(2)+v_(B)^(2)`......(ii) Both the above equation (i) and (ii) are plotted on `v_(A)-v_(B)` plane as shown in figure (ii). This plot can be used to find the unknowns `v_(A)` and `v_(B)`. For example the solution of the situation in figure (i) is `v_(A)=0,v_(B)=v_(0)` (point `y` in the plot) Because `v_(A)=v_(0), v_(B)=0` (point `x` in the plot) is not physically possible. In a situation block `A` is moving with velocity `2m//s` an strikes another identical block `B` kept at rest. The `v_(A)-v_(B)` plot for the situation is shown. `m` and `l` are the intersection points whose `v_(A), v_(B)` coordinaes are given in the figure. The coefficient of restitution of the collision is A. `1/2`B. `1//3`C. `1`D. Collision no possible |
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Answer» Correct Answer - B `e=(4/3-2/3)/2=1/3` |
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| 11. |
An electron is placed just in the middle between two long fixed line charges of charge density `+lambda` each. The wires are in the xy plane (do not consider gravity) A. The equilibrium of the electron will be stable along x-directionB. the equilibrium of the electron will be stable along y-directionC. the equilibrium of the electron will be unstable along y-directionD. the equilibrium of the electron will be stable along z-direction |
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Answer» Correct Answer - D If we displace the electron slightly towards X direction, it will thrown away towards right. so eq. is unstable along x direction. if we displace the electron slightly towards y direction, no Extra force will act. So eql. Is neutral along y axis . If we displace the electron towards z direction, it will be attract and try to come to eql. positron. so eql. is stable along z direction. |
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| 12. |
Graphical soluton of a two body head on collision A block `A` of mass `m` moving with a uniform velocity `v_(0)` strikes another identical block `B` kept at rest on a horizontal smooth surface as shown in the figure (i). We can conserve linear momentum. So `mv_(0)=mv_(A)mv_(B)` (`v_(A)` and `v_(B)` are the velocities of the blocks after collision) `:. v_(0)=v_(A)+v_(B)`.........(i) If the collision is perfectly elastic `1/2 mv_(0)^(2)=1/2 mv_(A)^(2)+1/2 mv_(B)^(2)` `impliesv_(0)^(2)=v_(A)^(2)+v_(B)^(2)`......(ii) Both the above equation (i) and (ii) are plotted on `v_(A)-v_(B)` plane as shown in figure (ii). This plot can be used to find the unknowns `v_(A)` and `v_(B)`. For example the solution of the situation in figure (i) is `v_(A)=0,v_(B)=v_(0)` (point `y` in the plot) Because `v_(A)=v_(0), v_(B)=0` (point `x` in the plot) is not physically possible. If the collision is perfectly inelastic, then the `v_(A)-v_(B)` plot isA. B. C. D. |
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Answer» Correct Answer - B For perfectly in elastic collision both blocks have same velocity after collision. |
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| 13. |
A uniform rope of their mass density `lamda` and length `l` is coiled on smooth horizontal surface. One end is pulled up by an external agent wit consant vertical velocity `v`. Choose the correct option(s) A. Power developed by external agent as a function of `x` is `P=lamdaxgv`B. Power developed by external agent as a function of `x` is `P=(lamdav^(2)+lamdaxg)v`C. Energy lost during the complete lift of the rope is zeroD. Energy lost during the complete lift of the rope is `(lamdalv^(2))/2` |
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Answer» Correct Answer - B::D `F` applied by external agent `=` Weight `+` thrust force `=lamdaxg+lamdav^(2)` Energy lost in the complete lift `=W_(F)-DeltaK.E. -DeltaU=(lamda lv^(2))/2` |
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| 14. |
A solid cylinder rolls from the back of a large truck travelling at `10 m//s` to the right. The cylinder is travelling horizontally at `8m//s` to the left relative to an observer in the truck. The ball lands on the roadway `1.25m` below its starting level. How far behind the truck does it land (in `m`)? |
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Answer» Correct Answer - D `1.25=1/2(10)t^(2)` `:.t=0.5` sec Distance of the cylinder from the truck `=8xx0.5=4m` |
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| 15. |
A body of mass `m` is moving along a circular path of radius `R` such that its kinetic energy at any instant is `k=k_(0)((t)/(t_(0)))^(2)` where `t_(0)` is an appropriate constant. ThenA. The magnitude of tangential component of force acting on it must be constant.B. The magnitude of centripetal force acting on the body will be directly proportional to `t^(2)`.C. After a long time, the resultant force will make a very small angle with the radius.D. The magnitude of centripetal force acting on the body will be directly proportional to `t`. |
| Answer» Use `F_(t)=m(dv)/(dt)` and `F_(t)=(mv^(2))/(r )` | |
| 16. |
A thin metal plate is being bombarded by a perpendicular beal of gas particles from both sides as shown in the figure. The solid dots are representing the melecules hitting from left side and the faint dots are the molecules hitting frm right side. The mass of these gas particles is `m=10^(-26)kg` and velocity before hitting is `v_(0)=5m//s`. Volume density of the gas particles on both sides is `n=10^(25)` per `m^(3)`. Each beam has an area `A=1m^(2)` and the collisions are perfectly elatic. What is the external force `F` (in newton) required to move the plate with a constant velocity `v=2m//s` |
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Answer» `F=2m eta A[(v_(0)+v)^(2)-(v_(0)-v)^(2)]` `=2(10^(-26))(10^(25))(1)[(5+2)^(2)-(5-2)^(2)]` `=0.2xx40=8N` |
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| 17. |
A truck has to move to a dimetrically opposite point on a circular track which surrounds a field. The speed of the truck along the track is `2v_(0)`. While that in the field is `v_(0)`. The driver plans to move along an arc of a circle and then along a straight line as shown. A. To reach `P` in shortest time, `Q` must be equal to `60^(@)`B. The minimum time required to rach `P` is `(R )/(v_(0))[(pi)/(6)+sqrt(3)]`C. The distance travelled to reach `P` in shortest time is `R[(pi)/(3)+sqrt(3)]`.D. The angle `theta` will not depend on the value of `v_(0)`. |
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Answer» `T=(Rtheta)/(2v_(0))+(2Rcos(theta//2))/(v_(0))` `(dT)/(d theta)=0rArr (R )/(2v_(0))-(Rsin(theta//2))/(v_(0))=0` `rArr sin(theta//2)=1//2rArrtheta=60^(@)rArrT=(R )/(v_(0))[(pi)/(6)+sqrt(3)]` Distance `=Rxx(pi)/(3)+2Rxx(sqrt(3))/(2)` |
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| 18. |
A very large plank `P` of some unknown mass is being moved with velocity `v_(0)hati` under application of an external force (not shown in figure). Simultaneously a block `B` of mass `m` placed on the plank is also moving with velocity `v_(0)hatk`. All these velocities are with respect to ground frame and at `t=0`. Coefficient of frictioin between the plank and the block `mu`. Choose the correct option(s). A. Kinetic friction force acting on the block at `t=0` is `-mumghatk`B. At `t=0`, power developed (with respect to ground frame) by kinetic friction force on the block `B` is `-(mu mg v_(0))/(sqrt(2))`C. At `t=0` power developed (with respect to ground frame) by kinetic friction force on the block `B` is `-mu mg sqrt(2) v_(0)`D. At `t=0`, heat dissipation per sec in the system is `2sqrt(2)mu mg v_(0)` |
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Answer» Correct Answer - B::C `vecf_(k)` on the block `B=mu mg(1/(sqrt(2)) hati-1/(sqrt(2))hatk)` `P_(f_(k))=mu mg [(hati)/(sqrt(2))-(hatk)/(sqrt(2))].(v_(0)hatk)` `=-(mu mg v_(0))/(sqrt(2))` `(dQ)/(dt)=mumgv_(0)2/(sqrt(2))=sqrt(2) mu mg v_(0)` |
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| 19. |
A semicircular wire of radius `R` is oriented vertically. A small bead is released from rest from the top of the wire. It slides without friction under the influence of gravity to the bottom, where it then leaves the wire horizontally and falls distance `H` to the ground. The bead lands a horizontal distance `D` away from where it was launched. Which of the following is correct graph of `RH` vetsus `D^(2)`? A. B. C. D. |
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Answer» Correct Answer - D `D=sqrt((2H)/(g)) sqrt(2g(2R))` |
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| 20. |
In given figure, a wire loop has been bent so that it has three segments ab (a quarter circle), bc (a square corner) & ca (straight line). Here are three choices for a magnetic field through the loop - (1) `vec_(B_(1))=3hati+7hatj-5thatk` (2) `vec(B_(2))=5thati-4hatj-15hatk` (3) `vec(B_(3))=2hati-5thatj-12hatk` where B is in milli tesla and t is in second. If the induced current in the loop due to `vec(B_(1)),vec(B_(2)),vec(B_(3))` are `i_(1), i_(2), i_(3)` respectively thenA. `i_(1)gti_(2)gti_(3)`B. `i_(2)gti_(1)gti_(3)`C. `i_(3)gti_(2)gti_(1)`D. `i_(1)=i_(2)=i_(3)` |
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Answer» Correct Answer - 2 `i_(1)=(dphi_(1))/(dt)(1)/(R)=(d[B_(1)xxArea])/(Rdt)` `=(d)/(dt)[5txx(pia^(2))/(4)]xx(1)/(R)=(5pia^(2))/(4R)` `i_(2)=(dphi_(2))/(dt)xx(1)/(R)=(d[5a^(2)t])/(dt)xx(1)/(R)=(5a^(2))/(R)` `i_(3)=(dphi_(3))/(dt)xx(1)/(R)=(d[0.5a^(2)])/(dt)xx(1)/(R)=(a^(2))/(R)` `thereforei_(2)gti_(1)gti_(3)` |
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| 21. |
A folded plate `OABCDEFO` made of materials such that part `OABO` (say part(i), `BCDOB` hs mass `4m, m` and `m` respectively. Part (i) is a uniform semicircular plate of radius `R//2` and in on the `xy` plane. Part (ii) and (iii) each is a uniform quarter circular plate of radius `R` on `xy` and `xz` plane respectively. The whole system is free to rotate about `y`-axis A particle `P` of mass `m` moving with velocity `v_(0)` hits to a point located at the circumcentre of the part (iii) and sticks to it. The point is at a distance `R//2` from `x` -axis as shown in the figure. Angular velocity of the combined system just after the collision is A. `(v_(0))/(4R)`B. `(2v_(0))/(5R)`C. `(3v_(0))/(5R)`D. `(v_(0))/(2R)` |
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Answer» Correct Answer - A Moment of inertia of combined system about `y` -axis `=(4m(R/2)^(2))/4+(mR^(2))/4+(mR^(2))/2+mR^(2)=2mR^(2)` Conservation of angular momentum along `y`-axis `mv_(0) R/2=2mR^(2)omega` `omega=(v_(0))/(4R)` |
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| 22. |
Four identical conducting rods are connected by pins. The velocity of point `R` is `v`. A uniform magnetic field `B` acts out of the paper. Choose the correct statement(s). (`PQ=PS=4l` and `QR=SR=3l`) A. The motional emf across `PQ` is `(6Blv)/(5)`B. The motional emf across `PQ` is `(3Blv)/(5)`C. The motional emf across `QR` is `(3Blv)/(5)`D. The motional emf across `QR` is `(6Blv)/(5)` |
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Answer» `v=20-x` Force `=mv(dv)/(dx)` |
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| 23. |
A moving company uses the pulley system in figure to lift heavy crates up a ramp. The ramp is coated with rollers that make the crate's motion essentially frictionless. A worker piles cinder blocks onto the plate until the plate moves down, pulling the crate up the ramp. Each cinder block has mass 10kg. The plate has mass 5kg. The rope is nearly massless, and the pulley is essentially frictionless. The ramp makes a 30° angle with the ground. the crate has mass 100kg.Let W1 denote the combined weight of the plate and the cinder blocks piled on the plate. Let T denote the tension in the rope. And let W2 denote the crate's weight.What is the smallest number of cinder blocks that need to be placed on the plate in order to lift the crate up the ramp? (A) 3 (B) 5 (C) 7 (D) 10 |
| Answer» The correct answer is (B) 5. | |
| 24. |
Particle running with constant speed v in circle of radius R then answer the following question. During running on circumference if particle made angle `theta` at centre then change in the magnitude of velocity.A. `2v sin. (theta)/(2)`B. zeroC. `2v cos. (theta)/(2)`D. `2v sin theta` |
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Answer» Correct Answer - B `|vec(v)_(2)|-|vec(v)_(1)|=0` |
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| 25. |
A uniform rod of length `4m` and mass `2sqrt(2)kg` revolves with constant angular velocity `omega` what a vertical axis through a smooth joint `A` at one extermly of the rod so that it describes a cone of semi vertical angle `53^(@)` as shown in the figure. Choose the correct statement (s). `(g=10m//s^(2))` A. The value of angular velocity `omega` is `2.5rad//s`B. The angle between hinge reaction at point `A` and vertical axis `AC` is `45^(@)`C. The magnitude of hinge reaction at point `A` is `40N`D. The angular momentum of the rod is constant. |
| Answer» `F=2xx(mu_(0)I_(1))/(2piR)xxI_(2)l=(2xx2xx10^(-7)xx50xx20xx0.50)/(0.25)=8xx10^(-4)N` | |
| 26. |
Particle running with constant speed v in circle of radius R then answer the following question. During running on circumferences if particle made angle `theta` at centre then magnitude of change in the velocityA. `2v sin. (theta)/(2)`B. zeroC. `2v cos. (theta)/(2)`D. `2v sin theta` |
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Answer» Correct Answer - A `|vec(v)_(2)-vec(v)_(1)|=2vsin.(theta)/(2)` |
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| 27. |
A particle of mass 0.40kg moving initially with a constant speed of 10ms-1 to the north is subject to a constant force of 8.0N directed towards the south for 30s. Take the instant the force is applied to be t = 0, the position of the particle at that time to be x = 0, and predict its position at t = - 5s, 25s, 100s. |
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Answer» a = - 20ms-2 0 ≤ t ≤ 30s t = - 5s: x = ut = -10 × 5 = - 50m t = 25s : x = ut + (½) at2 = (10 × 25 - 10 × 625)m = - 6km t = 100s : First consider motion up to 30s x1 = 10 × 30 -10 × 900 = - 8700m At t = 30s, v = 10 - 20 × 30 = - 590ms-1 For motion from 30s to 100s : x2 = 590 × 70 = - 41300m x = x1 + x2 = - 50km |
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| 28. |
A moving company uses the pulley system in figure to lift heavy crates up a ramp. The ramp is coated with rollers that make the crate's motion essentially frictionless. A worker piles cinder blocks onto the plate until the plate moves down, pulling the crate up the ramp. Each cinder block has mass 10kg. The plate has mass 5kg. The rope is nearly massless, and the pulley is essentially frictionless. The ramp makes a 30° angle with the ground. the crate has mass 100kg.Let W1 denote the combined weight of the plate and the cinder blocks piled on the plate. Let T denote the tension in the rope. And let W2 denote the crate's weight.The net force on the crate has magnitude: (A) W1 - W2sin300 (B) W1 - W2 (C) T - W2sin300 (D) T - W2 |
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Answer» (C) T - W2sin300 |
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| 29. |
A single wire `ACB` passes through a smooth ring at `C` when revolves at a constant speed in the horizontal circle of radius `r=6.4m` as shown in the figure. Find the speed (in `m//s)` of revolution of the ring. |
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Answer» Correct Answer - 8 Tension in both parts of the string will be the same `Tcos30^@+Tcos60^@=mg`…………i `Tcos30^@+Tcos60^@=mv^2//r`…………ii From eqn i and ii `v=sqrt(gr)=sqrt(10xx6.4)=8m//s` |
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| 30. |
Two forces are applied to a 5.0 kg crate, one is 6.0N to the north and the other is 8.0N to the west. The magnitude of the acceleration of the crate is :A. `0.50m//s^(2)`B. `2.0m//s^(2)`C. `2.8m//s^(2)`D. `10m//s^(2)` |
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Answer» Correct Answer - B `F_(x)=-8i" "F_(y)=6j` `|vec(F)|=10" "a=(F)/(m)=(10)/(5)=2m//s` |
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| 31. |
A cylindrical container of length `L` is full to the brim with a liquid which has mass density `rho`.it is placed on a weight -scale, the scale reading is `W`. A light ball of volume `V` and mass `m` which wold float on the liquid, if allowed to do so, is pushed gently downand held beneath the surface of the liquid with a rigid rod of negligible volume as shwon on the left. What is the mass `M` of the liquid which overflowed while the ball was being pushed into the liquid?A. `rhogV`B. `m`C. `mrho-V`D. none |
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Answer» Correct Answer - A Volume of the liquid that will flow out is equal to volume of the ball. So mass of the liquid that will overflow is `M=`volume overflow`xx`density of liquid `implies M=rhoV` |
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| 32. |
What is the length of projection of `vec(A)=3hat(i)+4hat(j)+5hat(k)` on `xy` plane ?A. `5`B. `3`C. `5sqrt2`D. `4` |
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Answer» Correct Answer - A Legnth of projection on `xy` plane is `=sqrt(3^(2)+4^(2))=5` |
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| 33. |
A particle moves along a straight line in such a way that its acceleration is increasing at the rate of `2 m//s^(3)`. Its initial acceleration and velocity were zero. Then, the distance which it will cover in the `3^(rd)` second `(t=2 " to "t =3 " sec")` is :A. `19//3 m`B. `12//5 m`C. `17//5 m`D. `19//4 m` |
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Answer» Correct Answer - A `(da)/(dt)=2impliesunderset(o)overset(a)intda=underset(o)overset(t)int2dtimplies a=2t` `a=(dv)/(dt)=2timpliesunderset(o)overset(v)intdv=underset(D)overset(t)int2tdtimpliesv=t^(2)` `v=(dx)/(dt)=t^(2)impliesunderset(o)overset(x)intdx=underset(2)overset(3)intt^(2)dt` `impliesx=(t^(3))/(3)|_(2)^(3)=(27)/(3)-(8)/(3)=(19)/(3)m` |
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| 34. |
The relationship between the displacement travelled by a body and the time t is described by the equation `s = A + Bt + Ct^(2) + Dt^(3)`. Where `C = 0.14m//s^(2)` and `D=0.01m//s^(3)`. In what time after motion begins will the acceleration of the body be equal to `1 m//s^(2)?`A. 10 secB. `50 // 3 "sec"`C. 12 secD. 18 sec |
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Answer» Correct Answer - C `S=A+Bt+Ct^(2)+Dt^(3)` `V=B+2Ct+3Dt^(2)` `a=2C+6Dt` `1=2xx0.14+6xx0.01t` t = 12 sec |
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| 35. |
All the blocks are attached to an ideal rope which passes over an ideal pulley. If acceleration of blocks `m_(1),m_(2),m_(3)` and `m_(4)` and `a_(1),a_(2),a_(3)` and `a_(4)` respectively then choose the correct option.A. `{:(a_(1),a_(2),a_(3),a_(4)),(guarr,2guarr,3guarr,g//suarr):}`B. `{:(a_(1),a_(2),a_(3),a_(4)),(guarr,2g//3darr,3guarr,g//2darr):}`C. `{:(a_(1),a_(2),a_(3),a_(4)),(2g//3darr,2guarr,3guarr,guarr):}`D. `{:(a_(1),a_(2),a_(3),a_(4)),(gdarr,2//gdarr,3gdarr,2g//3darr):}` |
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Answer» Correct Answer - B Case :I `T_(1)=2mg` `ma_(1)=2mg-T_(1)` `impliesa_(1)=guarr` Case II: `T_(2)=mg` `3ma_(2)=3mg-T_(2)` `impliesa_(2)=2g//3darr` Case III: `T_(3)=4mg` `ma_(3)=T_(3)-mg` `impliesa_(3)=3guarr` Case IV: `T_(4)=mg` `2ma_(4)=2mg-T_(4)` `impliesa_(4)=g//2darr` |
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| 36. |
Two blocks are arranged as shown in figure. Find the ratio of `a_(1)//a_(2)`. (`a_(1)` is acceleration of `m_(1)` and `a_(2)` that of `m_(2))` |
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Answer» Correct Answer - 6 Applying constraints : `a_1=6a_2` |
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| 37. |
When a `2` kg car driven at `20 m//s` on a level road is suddenly put into netural gear (i.e. allowed to coast), the velocity decreases in the following manner: `V=20/(1+(t/20))m//s` Where t is time in sec. The deceleration of car at the instant its speed is `10 m//s` is.A. `5` wattB. `10` wattC. `15` watt`D. `20` watt |
| Answer» Correct Answer - `F=m(dV)/(dt)V` | |
| 38. |
From a uniform disc of radius R, an equilibatered triangle of side `sqrt3R` is cut as shown. The new position of centre of mass is: A. `(0,0)`B. `(0,R)`C. `(0,(sqrt(3)R)/(2))`D. none of these |
| Answer» Centre of mass remain mass remain at centre of the circle. | |
| 39. |
The potential energy in given according to the following graph. Which of the following graph represent the force?A. B. C. D. |
| Answer» F=-Slope of U-x curve. | |
| 40. |
The elastic constant of the spring in figure is `25.00 N//m` and the mass of the block is `0.50` kg. If the block is to hit the bottle, what is minimum value of `v_(a)`. The spring has natural length when block is `x = 0 m`. A. `0.44 m//s`B. `0.88 m//s`C. `0.20 m//s`D. `0.15 m//s` |
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Answer» `W_(s)+W_(g)+W_(n)=Deltak` `1//2k 0.05^(2)-1//2K 0.08^(2)+0+0` `=0-1//2mv_(0)^(2)` |
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| 41. |
Which of the following does not exist-A. `PCl_(5)`B. `NCl_(3)`C. `NOCl_(3)`D. `NCl_(5)` |
| Answer» Due to absence of vacent d-orbital in Nitrogen and `NCl_(5)` is not exist. | |
| 42. |
The ratio of speeds of diffusion of two gases A and B is 1:4. If the mass ratio of A to B present in the given mixture is 2:1, then which of the following is the ratio mole-fraction of A to B?A. `2:3`B. `1:8`C. `2:1`D. `1:2` |
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Answer» `(r_(A))/(r_(B))=(1)/(4)=(n_(A))/(n_(B))sqrt((M_(B))/(M_(A)))` `(1)/(4)=(m_(A)//M_(A))/(m_(B)//M_(B))sqrt((M_(B))/(M_(A)))` `(1)/(4)=(m_(A))/(m_(B))((M_(B))/(M_(A)))^(3//2)` `(1)/(4)=(2)/(1)((M_(B))/(M_(A)))^(3//2)` `((M_(B))/(M_(A)))(8)^(2//3)=(1)/(4)` `(n_(A))/(n_(B))=(m_(A))/(m_(B))xx(M_(B))/(M_(A))=(2)/(1)xx(1)/(4)` =1:2 |
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| 43. |
A small object of mass m starts from rest at the position shown and slides along the fricationless loop-the -loop track of radius R. What is the smallest value of y such that the object will slide without losing contact with the track? A. `R//4`B. `R//2`C. `R`D. `2R` |
| Answer» `V=sqrt(gR)=sqrt(2gy)` | |
| 44. |
Select the pair of almost same size.A. Al,CaB. Zr,HfC. Fe,CoD. All of these |
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Answer» `Zr,Hfrarr` due to lanthanoid contraction `Al,Gararr` due to o-orbital contraction. |
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| 45. |
Diamond has face-centred cubic lattice. There are two atoms at `(0, 0, 0)` and `((a)/(4), (a)/(4), (a)/(4))` coordinates. The ratio of the carbon-carbon bond distance to the edge of the unit cell isA. `sqrt(3/16)`B. `sqrt(1/4)`C. `1/4`D. `1/sqrt2` |
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Answer» Correct Answer - A Carbon atoms are at corners and are at alternate corners.So from geometry `sqrt3(a/2)1/2=2r` So required ratio`=(2r)/a=sqrt3/4=sqrt(3/16)` |
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| 46. |
Which of the following statements is correct?A. Ionisation energy of `A^(-)` is greater than a when a is a halogen atom.B. Ionisation energy of `A^(+)` is greater than that of `A^(2+)` when A is the member of alkali metals.C. Successive ionisation energy is always increasing for `1^(st)` and `2^(nd)` period element.D. Electron affinity value of `A^(+)` is numerically identical with the ionisation potential of `A^(-)` [for any atom]. |
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Answer» (I) Inosation energy of A is less than A (2) Ionisation energy of `A^(+)` is less than that of `A^(2+)` |
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| 47. |
Select the ion which has inert gas configuration but follow the octer rule-A. `B^(+3)`B. `Al^(+3)`C. `Ga^(+3)`D. `Ge^(+4)` |
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Answer» `B^(+3)rarrHe` `Al^(+3)rarrNe` `underset(Ge^(+4))overset(Ge^(+3)).}rarrns^(2),np^(6),nd^(10)` pseudo inert gas configuration |
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| 48. |
Which of the following Lewis structure is not valied for Azide ion `(N_(3)^(-))` ?A. `:N-=N-underset(..)overset(..)N:^(2-)`B. `:N-=overset(+)N-underset(..)overset(..)N:^(2-)`C. `:underset(..)overset(-)N-=overset(+)N=underset(..)overset(-)N:`D. None of these |
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Answer» Correct Answer - B::D For Zzide ion `n_(1)=3xx5+1=16e^(-)` (in valence shell) `n_(2)=3xx8=24` `n_(3)=24-16=8,bpr =(8)/(2)=4` `n_(4)=16-8=8` inpr `=(8)/(2)=4` In 1 st Lewis structure of Azide ion - F.C. on N (a)=5-3-2=0(correct) `:N-=N-underset(..)overset(..2)N` F.C. on N(b)=5-4=+1 (not shown in option) F.C. on N(c)=5-1-6=-2(correct) So 1st Lewis structure show wrong representation of formal charge on N(b). |
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| 49. |
The suffix of principal group, the prefixes for the other groups and the name of the parent in the structure. `HO-CH_(2)-underset(CH_(3))underset(|)CH-CH=CH_(2)-underset(O)underset(||)C-underset(O)underset(||)C-NH_(2)`A. amide, hydroxy, amino, formyl, methyl, hept-4eneB. one, carbanoyl, amino, hydroxy, methyl,oxo, hopt-4-eneC. amide, amino, hydroxy, methyl,oxo hept-4-eneD. amine, carbanoyl, hydroxy, methyl,oxo,hept-4-ene |
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Answer» Principal group`rarr-underset(O)underset(||)C-NH_(2)` Other group `rarr-NH_(2),-underset(O)underset(||)C-,-OH` Name of parent chain `rarr` Heptane Suffix `rarr` ene |
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| 50. |
State the factors on which the refractive index of a material medium for a given wavelength depends. |
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Answer» Refractive index is a characteristic property of the medium, whose value depends only on nature of material of the medium and the color or wavelength of light. Also, Depends on temperature and nature of surrounding medium. |
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