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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.
| 1. |
A particle moves on a line according to the law `s=at^(2)+bt+c`. If the displacement after 1 sec is 16 cm, the velocity after 2 sec is 24 cm/sec and acceleration is `8cm//sec^(2)`, thenA. a=4, b=8, c=4B. a=4, b=4, c=8C. a=8, b=4, c=4D. none of these |
| Answer» Correct Answer - A | |
| 2. |
A particle moves along the parabola `y^2=2ax` in such a way that its projection on y-axis has a constant velocity. Show that its projection on the x-axis moves with constant accelerationA. constant velocityB. constant accelerationC. variable velocityD. variable acceleration |
| Answer» Correct Answer - B | |
| 3. |
A spherical balloon is being inflated so that its volume increase uniformaly at the rate of `40 cm^(3)//"minute"`. The rate of increase in its surface area when the radius is 8 cm, isA. `10 cm^(2)` minuteB. `20 cm^(2)` minuteC. `40 cm^(2)` minuteD. none of these |
| Answer» Correct Answer - A | |
| 4. |
The edge of a cube is equal to the radius o the sphere. If the rate at which the volume of the cube is increasing is equal to `lambda`, then the rate of increase of volume of the sphere, isA. `(4pi lambda)/(3)`B. `4pi lambda`C. `(lambda)/(3)`D. none of these |
| Answer» Correct Answer - A | |
| 5. |
At an instant the diagonal of a square is increasing at the rate of 0.2 cm/sec and the area is increasing at the rate of `6 cm^(2)//sec`. At the moment its side, isA. `(30)/(sqrt(2))cm`B. `30sqrt(2)` cmC. 30 cmD. 15 cm |
| Answer» Correct Answer - A | |
| 6. |
The weight W of a certain stock of fish is given by W = nw, where n is the size of stock and w is the average weight of a fish. If n and w change with time t as n = `2t^(2)+3` and `w=t^(2)-t+2`, then the rate of change of W with respect to t at t = 1, isA. 1B. 13C. 5D. 8 |
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Answer» Correct Answer - B We have, `W=nw, n=2t^(2)+3 and w=t^(2)-t+2` `therefore (dV)/(dt)=(dn)/(dt)w+n(dw)/(dt),(dn)/(dt)=4t,(dW)/(dt)=2t-1` At t=1, we get `n=5,w=2, (dn)/(dt)=4, (dw)/(dt)=1` Hence, `((dW)/(dt))_(t=1)=4xx2+5xx1=13` |
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| 7. |
Side of an equilateral triangle expands at therate of 2 cm/sec. The rate of increase of its area when each side is 10 cm is(a)cm2/sec (b)cm2/sec(c) 10 cm2/sec (d) 5 cm2/secA. `10sqrt(2)cm^(2)//sec`B. `10sqrt(3)cm^(2)//sec`C. `10 cm^(2)//sec`D. `5cm^(2)//sec` |
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Answer» Correct Answer - B Let x be the side and A be the area of equilateral triangle at time t. Then, `A=(sqrt(3))/(4)x^(2)` `implies(dA)/(dt)=sqrt(3)/(2)xx(dx)/(dt)` `implies(dA)/(dt)=(sqrt(3))/(2)xx10xx2=10sqrt(3)" "[because x=10 and (dx)/(dt)=2]` |
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| 8. |
The surface area of a cube is increasing at the rate of `2 cm^(2)//sec`. When its edge is 90 cm, the volume is increasing at the rate ofA. `1620 cm^(3)//sec`B. `810 cm^(3)//sec`C. `405 cm^(3)//sec`D. `45 cm^(3)//sec` |
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Answer» Correct Answer - D Let at any time t the length of each edge of the cube be x cm. Let S denote the surface area and V the volume of the cube. Then, `S=6x^(2) and V=x^(3)` `implies (dS)/(dt)=12x (dx)/(dt) and (dV)/(dt)=3x^(2)(dx)/(dt)` `implies 2=12 xx 90(dx)/(dt) and (dV)/(dt)=3xx90^(2)xx(dx)/(dt) [because x=90 cm and (dS)/(dt)=2]` `implies 90xx(dx)/(dt)=(1)/(6) and (dV)/(dt)=3xx90xx(90xx(dx)/(dt))` `implies (dV)/(dt)=3xx90xx(1)/(6) cm^(3)//sec=45 cm^(3)//sec` |
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| 9. |
A spherical iron ball 10cm in radius is coated with a layer of ice ofuniform thickness that melts at a rate of `50c m^3//m in`. When the thickness of ice is 5cm, then findthe rate at which the thickness of ice decreases.A. `(5)/(6pi)cm//min`B. `(1)/(54pi) cm//min`C. `(1)/(18pi) cm//min`D. `(1)/(36pi) cm//min` |
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Answer» Correct Answer - C Let at any time, h cm be the thickness of ice. Then, V=Volume of ice `=(4)/(3)pi(10+h)^(3)-(4)/(3)pixx10^(3)` `implies (dV)/(dt)=4pi(10+h)^(2)(dh)/(dt)` `implies -50=4pi (10+5)^(2)xx(dh)/(dt) " "[because (dv)/(dt)=-50 cm^(3)//min and h = 5]` `implies (dh)/(dt) = -(1)/(18pi) cm//min` |
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| 10. |
Find the surface area of a sphere when its volumeis changing at the same rate as its radius.A. 1B. `(1)/(2sqrt(pi))`C. `4pi`D. `(4pi)/(3)` |
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Answer» Correct Answer - A At any time t, let V, S and r denote respectively the volume, surface area and radius of the sphere. Then, `V=(4)/(3)pir^(3) and S=4pir^(2)` `therefore (dV)/(dt)=(dr)/(dt)implies4pir^(2)(dr)/(dt)=(dr)/(dt)implies4pir^(2)=1impliesS=1` |
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| 11. |
The rate of change of surface area of a sphere of radius r when the radius is increasing at the rate of 2 cm/sec is proportional toA. `(1)/(r^(2))`B. `(1)/(r )`C. `r^(2)`D. r |
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Answer» Correct Answer - D Let S be the surface area and r be the radius of the sphere at any time t. Then, `S=4pi r^(2)` `implies (dS)/(dt)=8pi r(dr)/(dt)` `implies (dS)/(dt)=8pirxx2" "[(dr)/(dt)=2cm//sec]` `implies (dS)/(dt)=16pir implies (dS)/(dt)propr` |
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| 12. |
A cylindrical vessel of radius 0.5 m is filledwith oil at the rate of 0.25m3/minute. The rate at which thesurface of the oil is rising, is(a) 1 m/min.(b) 2 m/min. (c) 5 m/min. (d) 1.25 m/min.A. 1 m / minuteB. 2 m / minuteC. 5 m / minuteD. 1.25 m / minute |
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Answer» Correct Answer - A Let r be the radius, h be the height and V be the volume of the oil at time t. Then, `V=pir^(2)h` `impliesV=(pi)/(4)h" "[because r=0.5 m =(1)/(2)"(given)"]` `implies (dV)/(dt)=(pi)/(4)(dh)/(dt)` `implies 0.25 pi=(pi)/(4)xx(dh)/(dt) " "[because(dV)/(dt)=0.25 pi m^(3) "(given)"]` `implies (dh)/(dt)=1m//"minute"` |
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| 13. |
For what values of x is the rate of increase of `x^(3)-5x^(2)+5x+8` is twice the rate of increase of x?A. `-3, -(1)/(3)`B. `-3, (1)/(3)`C. `3, -(1)/(3)`D. `3, (1)/(3)` |
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Answer» Correct Answer - D Let `y=x^(3)-5x^(2)+5x+8`. Then, `(dy)/(dt)=(3x^(2)-10x+5)(dx)/(dt)` When `(dy)/(dt)=2(dx)/(dt)`, we have `(3x^(2)-10x+5)(dx)/(dt)=2(dx)/(dt)` `implies 3x^(2)-10x+3=0` `implies(3x-1)(x-3)=0` `impliesx=3,(1)/(3)` |
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| 14. |
If the rate of change of sine of an angle `theta` is k, then the rate of change of its tangent isA. `k^(2)`B. `(1)/(k^(2))`C. kD. `(1)/(k)` |
| Answer» Correct Answer - B | |
| 15. |
If the rate of change of area of a square plate is equal to that of the rate of change of its perimeter, then length of the side, isA. 1 unitB. 2 unitsC. 3 unitsD. 4 units |
| Answer» Correct Answer - B | |
| 16. |
A cone whose height always equal to its diameter is increasing in volume at the ran `40 (cm^3)/sec` .At what rate is the radius increasing when its circular base area is `1m^2?` (A) 1 (B)0.001 (C) 2 (D) 0.002A. 1 mm/secB. 0.001 cm/secC. 2 mm/secD. 0.002 cm/sec |
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Answer» Correct Answer - D Let h be the height, r be the radius of the base and V be the volume of the cone at time t. Then, `V=(1)/(3)pir^(2)h` `implies V=(2)/(3)pi r^(3)" "[because h = 2r]` `implies (dV)/(dt)=2pir^(2)(dr)/(dt)` `implies 40=2(10^(4))(dr)/(dt)" "[because pir^(2)=1m^(2)" (given)"=10^(4) cm^(2) and (dV)/(dt)=40cm^(3)//sec]` |
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| 17. |
If the semivertical angle of a cone is `45^@`. Then the rate of change of its volume isA. curved surface area times the rate of change of rB. base area times the rate of change of lC. base area times the rate of change of rD. none of these |
| Answer» Correct Answer - B | |
| 18. |
The points of the ellipse `16x^2+9y^2=400` at which the ordinate decreases at the same rate at which the abcissa increases isA. `(3, 16//3)`B. `(-3, 16//3)`C. `(3, 16//3)`D. `(3, -3)` |
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Answer» Correct Answer - A We have, `16x^(2)+9y^(2)=400` `implies 32x(dx)/(dt)+18 y (dy)/(dt)=0 " [Differentiating w.r.t. t]"` `implies 32x(dx)/(dt)-18y(dx)/(dt)=0 " "[because-(dy)/(dt)=(dx)/(dt)("given")]` `implies 16x-9y=0` `impliesy=(16x)/(9)` `therefore 16x^(2)+9y^(2)=400 implies 16x^(2)+(256)/(9)x^(2)=400impliesx=pm3` Now, `y=(16)/(9)x and x=pm3impliesy= pm(16)/(3)` Hence, the required points are `(3, 16//3)` and `(-3, -16//3)` |
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| 19. |
If the velocity v of a particle moving along a straight line and its distance s from a fixed point on the line are related by `v^(2)=a^(2)+s^(2)`, then its acceleration equalsA. asB. sC. `s^(2)`D. 2s |
| Answer» Correct Answer - B | |
| 20. |
The edge of a cube is equal to the radius of a sphere. If the edge and the radius increase at the same rate, then the ratio of the increases in surface areas of the cube and sphere isA. `2pi : 3`B. `3 : 2pi`C. `6 : pi`D. none of these |
| Answer» Correct Answer - B | |
| 21. |
The side of a square is equal to the diameter of a circle. If the side and radius change at the same rate, then the ratio of the change of their areas, isA. `1 : pi`B. `pi : 1`C. `2 : pi`D. `1 : 2` |
| Answer» Correct Answer - C | |
| 22. |
A variable `DeltaABC` is inscribed in a circle of diameter `x` units. At a particular instant the rate of change of side `a`, is `x/2` times the rate of change of the opposite angle `A` then `A =`A. `pi//6`B. `pi//3`C. `pi//4`D. `pi//2` |
| Answer» Correct Answer - B | |
| 23. |
The radius of a sphere is changing at the rate of0.1 cm/sec. The rate of change of its surface area when the radius is 200 cmis(a)cm2/sec (b)cm2/sec(c)cm2/sec (d)cm2/secA. `8pi cm^(2)//sec`B. `12pi cm^(2)//sec`C. `160pi cm^(2)//sec`D. `200cm^(2)//sec` |
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Answer» Correct Answer - C Let r be the radius and S be the surface area of the sphere at any time t. Then, `S=4pir^(2)` `implies (dS)/(dt)=8pir(dr)/(dt)` `implies ((dS)/(dt))_(r=200)=8pixx200xx0.1=160 pi cm^(2)//sec` |
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| 24. |
The radius and height of a cylinder are equal. If the radius of the sphere is equal to the height of the cylinder, then the ratio of the rates of increase of the volume of the sphere and the volume of the cylinder, isA. `4 : 3`B. `3 : 4`C. `4 : 3pi`D. `3pi : 4` |
| Answer» Correct Answer - A | |
| 25. |
The radius of the base of a cone is increasing at the rate of 3 cm/minand the altitude is decreasing at the rate of 4 cm/min. The rate of change oflateral surface when the radius is 7 cm and altitude is 24cm is`108pic m^2//min`(b) `7pic m^2//min``27pic m^2//min`(d) none of theseA. `54pi cm^(2)//min`B. `7 pi cm^(2)//min`C. `27 cm^(2)//min`D. none of these |
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Answer» Correct Answer - A Let r, l and h denote respectively the radius, slant height and height of the cone at any time l. Then, `l^(2)=r^(2)+h^(2)` `implies 2l(dl)/(dt)=2r(dr)/(dt)+2h(dh)/(dt)` `implies l(dl)/(dt)=r(dr)/(dt)+h(dh)/(dt)` `l(dl)/(dt)=7xx3+24xx-4 " "[because(dh)/(dt)=-4and (dr)/(dt)=3]` `implies l(dl)/(dt)=-75` When r = 7 and h=24, we have `l^(2)=7^(2)+24^(2)" "[because l^(2)=r^(2)+h^(2)]` implies l = 25 `l(dl)/(dt)=-75implies(dl)/(dt)=-3` Let S denote the laternal surface area. Then, `S=pi r l` `implies (dS)/(dt)=pi{(dr)/(dt)l+r(dl)/(dt)}=pi{3xx25+7xx-3}=54pi` |
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| 26. |
If the length of the diagonal of a square is increasing at the rate of 0.2 cm/sec, then the rate of increase of its area when its side is `30//sqrt(2)` cm, isA. `3 cm^(2)//sec`B. `(6)/(sqrt(2))cm^(2)//sec`C. `3sqrt(2)cm^(2)//sec`D. `6cm^(2)//sec` |
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Answer» Correct Answer - D Let at any time t the length of the diagonal be x cm. Then, each side = `(x)/(sqrt(2))` cm. We have, `(dx)/(dt) = 0.2 cm//sec` Let A be the area of the squar. Then, `A=((x)/(sqrt(2)))^(2)=(1)/(2)x^(2)` `implies (dA)/(dt)=x(dx)/(dt)` `implies ((dA)/(dt))_((x)/(sqrt(2))=(30)/(sqrt(2)))=30xx0.2 = 6cm^(2)//sec` |
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| 27. |
If `V=(4)/(3)pir^(3)`, at what rate in cubic units is V increasing when r = 10 and `(dr)/(dt)` = 0.01?A. `pi`B. `4pi`C. `40pi`D. `4pi//3` |
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Answer» Correct Answer - B We have, `V=(4)/(3) pir^(3)` `implies(dV)/(dt)=4pir^(2)(dr)/(dt)` `implies (dV)/(dt)=4pixx100xx0.01 " "[because r = 10 and (dr)/(dt)=0.01]` `implies (dV)/(dt)=4pi` Hence, V is increasing at the rate of `4pi` cubic units per unit of time. |
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| 28. |
Gas is being pumped into a a spherical balloon at the rate of `30 ft^(3)//min`. Then the rate at which the radius increases when it reaches the value 15 ft, isA. `(1)/(30pi) ft//min`B. `(1)/(15pi)ft//min`C. `(1)/(20)ft//min`D. `(1)/(25)ft//min` |
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Answer» Correct Answer - A Let r be the radius of the spherical balloon and V be the volume at any time t. It given that `(dV)/(dt) = 30 ft^(3)//min` Now, `V=(4)/(3) pi r^(3)` `implies (dV)/(dt)=4 pi r^(2)=(dr)/(dt)` `implies ((dV)/(dt))_(r=15) = 4pi xx 15^(2)xx((dr)/(dt))_(r=15)` `implies 30 = 900pi ((dr)/(dt))_(r=15)` `implies ((dr)/(dt))_(r=15)=(1)/(30pi) ft//min` |
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| 29. |
If `s=4t+(1)/(t)` is the equation o motion of a particle, then the acceleration when velocity vanishes, is |
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Answer» Correct Answer - B We have, `s=4t+(1)/(t)implies(ds)/(dt)=4-(1)/(t^(2))and (d^(2)s)/(dt^(2))=(2)/(t^(3))` Now, Velocity = `0 implies (ds)/(dt)=0implies 4-(1)/(t^(2))=0impliest=(1)/(2)` `therefore ("Acceleration at t ="(1)/(2))=((d^(2)s)/(dt^(2)))_(t=(1)/(2))=(2)/(((1)/(2))^(3))=16` |
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| 30. |
If a particle is moving such that the velocity acquired is proportional to the square root of the distance covered, then its acceleration isA. a constantB. `prop s^(2)`C. `prop(1)/(s^(2))`D. `prop s` |
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Answer» Correct Answer - A Let v be the velocity of the particle when the distance covered is s. Then, `v prop sqrt(s)` [Given] `implies v = lambda sqrt(s)` `implies (dv)/(ds)=(lambda)/(2sqrt(s))` `implies v(dv)/(ds)=(lambdav)/(2sqrt(s)) = (lambda^(2))/(2)` = constant. Hence, the acceleration is constant. |
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| 31. |
A particle moves in a straight line so that `s=sqrt(t)`, then its acceleration is proportional toA. `("velocity")^(3)`B. velocityC. `("velocity")^(2)`D. `("velocity")3//2` |
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Answer» Correct Answer - A We have, `s=sqrt(t)implies(ds)/(dt)=(1)/(2sqrt(t))and (d^(2)s)/(dt^(2))=(1)/(4t^(3//2))` `implies (d^(2)s)/(dt^(2))=-((2ds)/(dt))^(3)=-2((ds)/(dt))^(3)` Hence, Acceleration `prop` `("Velocity")^(3)`. |
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| 32. |
If the velocity of a body moving in a straight line is proportional to the square root of the distance traversed, then it moves withA. variable forceB. constant forceC. zero forceD. zero acceleration |
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Answer» Correct Answer - B Let v be the velocity and s be the distance traversed by the body at any time t. then, `v prop sqrt(s)` `implies v= lambda sqrt(s)`, where `lambda` is a constant `implies (dv)/(dt)=(lambda)/(2sqrt(s)) (ds)/(dt)` `implies (dv)/(dt)=(lambda)/(2sqrt(s)) (lambda sqrt(s))=(lambda^(2))/(2)`=const. implies Acceleration = Constant |
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| 33. |
If `s=ae^(l) + be^(-t)` is the equation of motion of a particle, then its acceleration is equal toA. sB. 2sC. 3sD. 4s |
| Answer» Correct Answer - A | |
| 34. |
If `s=e^(t)` (sin t - cos t) is the equation of motion of a moving particle, then acceleration at time t is given byA. `2 e^(t)` (cos t+sint)B. `2 e^(t)` (cos t-sint)C. `e^(t)` (cos t-sint)D. `e^(t)` (cos t+sint) |
| Answer» Correct Answer - A | |
| 35. |
If a point is moving in a line so that its velocity at time t is proportional to the square of the distance covered, then its acceleration at time t varies asA. cube of the distanceB. the distanceC. square of the distanceD. none of these |
| Answer» Correct Answer - A | |
| 36. |
The distances moved by a particle in time t seconds is given by `s=t^(3)-6t^(2)-15t+12`. The velocity of the particle when acceleration becomes zero, isA. 15B. `-27`C. `6//5`D. none of these |
| Answer» Correct Answer - B | |