InterviewSolution
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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.
| 51. |
Find the volume of the cone of height 10 cm and radius of the base 3 cm. |
| Answer» Correct Answer - `30 pi cm^3` | |
| 52. |
Find the (i) surface area and (ii) volume of a shpere of radius 5cm. |
| Answer» Correct Answer - (i) `100 pi cm^2` (ii) `166.7 pi cm^3` | |
| 53. |
A metallic disc is being heated. Its area A `("in" m^2)` at any time t `("in" sec)` is given by `A =5 t^2 +4 t+8` Calculate the rate of increase of area at `t =3 s`. |
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Answer» Correct Answer - `34 m^2//s` `(dA)/(dt)=(d)/(dt)(5t^2 +4t +8) = 10t+4` when `t =3 s , (dA)/(dt) =10xx3+4 =34m^2//s` |
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| 54. |
A particle starts from origin with uniform acceleration. Its displacement after t seconds is given in metres by the relation `x = 2 + 5 t + 7t^2` Calculate the magnitude of its (i) initial velocity (ii) velocity at t =4 s (iii) uniform acceleration (iv) displacement at t =5 s. |
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Answer» Correct Answer - (i) `5m//s` (ii) `61m//s` (iii) `14m//s^2` (iv) 202 m Velocity, `upsilon = (dx)/(dt) = (d)/(dt) (2+5t +7t^2) =(5 +14t)` (i) When `t =0, upsilon = 5m//s` (ii) When `t =4 s, upsilon =5 +14 xx4 =61m//s` (iii) Acceleration, `a=(dv)/(dt) = (d)/(dT) (5+14t) = 14m//s^2` (iv) When `t =5s, x =2 +5xx5+7xx5^2 = 2 +25 +175 =202m` |
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| 55. |
If the displacement x of a particle (in metre) is related with time (in second) according to relation `x = 2 t^3 - 3t^2 +2 t + 2` find the position, velocity and acceleration of a particle at the end of 2 seconds. |
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Answer» Correct Answer - `10 m ; 14 m//s ; 18 m//s^2` `x = 2t^3 - 3t^2 +2 t +2` velocity `=(dx)/(dt) = 6t^2 - 6t +2. "Acceleration" =(d)/(dT)((dx)/(dt)) = 12t -6` When `t = 2 sec, x =2xx2^3 - 3xx2^2 +2xx2+2 =16 -12 +4 +2 =10m` Velocity ` =6xx2^2 - 6xx2+2 =14m//s,` Acceleration `=12xx2 -6 =18m//s^2` |
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| 56. |
if the displacement of the particle at an instant is given by `y = r sin (omega t - theta)` where r is amplitude of oscillation. `omega` is the angular velocity and `-theta` is the initial phase of the particle, then find the particle velocity and particle acceleration. |
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Answer» Correct Answer - `r omega cos (omega t - theta) ; - omega^2 y` `y =r sin (omegat = theta)` Velocity, `upsilon = (dy)/(dt) =r cos (omega t -theta)(d)/(dt)(omega t - theta) = r cos (omegat - theta)xx omega =r omega cos (omegat - theta)` Acceleration, `A =(d upsilon)/(dt) = r omega [-sin(omegat - theta)]xxomega=- omega^2 r sin (omegat - theta) =- omega^2 y` |
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