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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.
| 151. |
State which of the following situations are possible and give an example for each of these;a) an object with a constant acceleration but with zero velocity.b) an object moving in a certain direction with an acceleration in the perpendicular direction. |
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Answer» a) This is possible b) This is possible |
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| 152. |
Write true or false for the following statements:A motion is said to be uniform if x ∝ t2 |
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Answer» False A motion is said to be uniform if velocity is constant. Velocity is constant means velocity does not change with time. Now, we know that, Displacement =\(x\propto\,time^2\) ….1 Also, we know that, Velocity = \(\frac{Displacement}{Time}\) ….2 From equations 1 and 2, we get, Velocity ∝ Time So, velocity increases with time and hence not constant. .So, if x ∝ time2 , then motion is not uniform. |
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| 153. |
When the motion of the body is said to be uniform? |
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Answer» The motion of the body is said to be uniform when its velocity is constant. |
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| 154. |
Observe the table given below answer the question:Time (t) 01234Distance (s)0481216What do you generalise from the table given above ? A) Non-uniform velocity B) Rest C) Accelerated motion D) Uniform velocity |
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Answer» D) Uniform velocity |
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| 155. |
Does a body have acceleration in the following situations? Why?body travelling along a straight line with uniform velocity.body travelling along a straight line with non-uniform velocity.body travelling along a circular path with uniform speed.body travelling along a circular path with non-uniform speed. |
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If an object should have acceleration it should be travelled in non uniform speed or in non uniform velocity. Change in velocity time Acceleration = \(\frac{change\,in\,velocity}{time}\) |
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| 156. |
The motion is said to be ……………….. if acceleration is constant. A) uniform motion B) uniform accelerated motion C) non-uniform accelerated motion D) none of these |
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Answer» B) uniform accelerated motion |
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| 157. |
Direction of acceleration is along the direction of ……………. A) velocity B) constant velocity C) increase in velocity D) all the above |
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Answer» Correct option is A) velocity |
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| 158. |
When a train comes to rest, then the acceleration is …………….. A) positive B) negative C) maximum D) none of these |
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Answer» Correct option is B) negative |
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| 159. |
A car covers half the distance at a speed of 60 kmh-1 and the other half at a speed of 40 km h-1 The average speed of the car is A) 44.44 km h-1 B) 50 km h-1 C) 48 km h-1 D) 0 |
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Answer» Correct option is B) 50 km h-1 |
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| 160. |
A man is sitting on a train which is moving. Is he at rest or in motion? |
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Answer» In Classical Mechanics, we deal with relative motion, that is, we study motion with respect to an observer. So, first thing that we need to do is to set an observer and study motion with respect to this observer which is at rest. (a) A man sitting in a train is at rest with respect to a man sitting in the train next to him. (b) A man sitting in a train is in motion with respect to a man standing on ground outside the train, because the train is moving with respect to the observer and so is the man sitting in it. |
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| 161. |
An electric engine has a velocity of 120 kmh-1. How much distance will it travel in 30s? |
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Answer» Now, | velocity l =120 kmh-1 = \(\frac{120\times1000}{2600} ms^{-1}\) \(= \frac{100}{3}ms^{-1}\) Now, time = 30 s Hence, Distance = |velocity| \(\times\) time \(=\frac{100}{3}\times30\) = 1000 m = 1 km Hence the distance travelled by the electric engine in 30 s is 1 km. |
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| 162. |
An aircraft travelling at 600 km/h accelerates steadily at 10 km/h per second. Taking the speed of sound as 1100 km/h at the aircraft’s altitude, |
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Answer» Initial velocity, u = 600 km/h Final velocity, v = 1100 km/h Acceleration = 10 km/h/s = 600 km/h2 From relation, a = (v-u)/t t = (v-u)/a t = (1100-600)/600 = 500/600 = 5/6 hr = 50 sec |
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| 163. |
An aircraft travelling at 600 km/h accelerates steadily at 10 km/h per second. Taking the speed of sound as 1100 km/h at the aircraft’s altitude how long will it take to reach the ‘sound barrier’? |
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Answer» Initial velocity, u = 600 km/h Final velocity, v = 1100 km/h Acceleration = 10 km/h/s = 600 km/h2 From formula, a = (v-u)/t t = (v-u)/a t = 50 sec |
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| 164. |
A train travelling at 20 m/s accelerates at 0.5 m/s2 for 30s. How far will it travel in this time? |
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Answer» Initial velocity, u = 20 m/s Time, t = 30s Acceleration, a = 0.5 m/s2 s = ut = ½ at2= 825m |
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| 165. |
A cyclist is travelling at 15 m/s. She applies brakes so that she does not collide with a wall 18 m away. What deceleration must she have? |
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Answer» Initial velocity, u = 15 m/s Final velocity, v = 0 m/s Distance, s = 18m Acceleration, a = ? Deceleration, v2 – u 2= 2as = 6.25 m/s2 |
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| 166. |
An aircraft travelling at 600 km/h accelerates steadily at 10 km/h per second. Taking the speed of sound as 1 |
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Answer» Initial velocity, u = 600 km/h Final velocity, v = 1100 km/h Acceleration = 10 km/h/s = 600 km/h2 From formula, a = (v-u)/t t = (v-u)/a t = 50 sec |
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| 167. |
If a bus travelling at 20 m/s is subjected to a steady deceleration of 5 m/s2 , how long will it take to come to rest? |
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Answer» Deceleration, a = -5 m/s2 Initial velocity, u = 20 m/s Final velocity, v = 0 m/s Time, t = ? a = (v-u)/t -5 = (0-20)/t t = 20/5 = 4s |
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| 168. |
A bus is travelling the first one-third distance at a speed of 10 km/h, the next one-fourth at 20 km/h and the remaining at 40 km/h. What is the average speed of the bus? A. 17 km/h B. 17.8 km/h C. 18 km/h D. 20 km/h |
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Answer» Let the total distance be 100 km One third of 100 km = 33.33 km time = t1 =\(\frac{33.33}{10}\) = 3.33 hr Distance left = 100 km – 33.33 km = 66.67 km One fourth of 66.67 km = 16.67 km t2 =\(\frac{16.67}{20}\) = 0.83 hr Remaining distance = 100 km – 33.33 km – 16.67 km = 50 km t3 = \(\frac{50}{40}\)= 1.25 hr t1+t2+t3 = 5.41 hr Average speed = \(\frac{total\,distance\,travelled}{total\,time\,taken}\)= \(\frac{100}{5.41}\)= 18.4 km/hr |
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| 169. |
A boy begins to walk eastward along a street in front of his house and the graph of his position from have is shown in the following figure. His average speed for whole time interval in equal to A. 8 m/min B. 6 m/min C. \(\frac83\)m/minD. 2 m/min |
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Answer» Average speed = \(\frac{(total\,distance\,travelled)}{total\,time\,taken}\)= \(\frac{120}{20}\)m/min = 6m/min |
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| 170. |
You are sitting in a stationary car. There is a helium balloon tied to its floor. You accelerate and obviously feel like you are being pushed backwards (against the direction of your accelerations). The balloon. A. will move forward B. will move backward C. will remain state D. None of these |
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Answer» Helium is less denser than the air and so it is lighter then the air, so when the car is accelerated the air starts moving backwards but as the helium is lighter it is pushed forward by the air collecting at its back. |
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| 171. |
What does the slope of a speed-time graph indicate ?A. the distanceB. the displacementC. the accelerationD. the speed |
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Answer» Correct Answer - ( c) The slope of velocity time graph represents the acceleration. |
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| 172. |
What does the slope of a speed-time graph indicate ?A. distance travelledB. velocityC. accelerationD. displacement |
| Answer» Correct Answer - C | |
| 173. |
STATEMENT -1 : For an observer looking out through the window of a fast moving train , the nearby objects appear to move in the opposite direction to the train , while the distant objects appear to be stationary . STATEMENT - 2 : If the observer and the object are moving at velocities ` vec v_(1)` and `vec v_(2)` respecttively with refrence to a laboratory frame , the velocity of the object with respect to a laboratory frame , the velocity of the object with respect to the observer is `vecv_(2) - vecv(1) ` . (a) Statement -1 is True, statement -2 is true , statement -2 is a correct explanation for statement -1 (b) Statement 1 is True , Statement -2 is True , statement -2 is NOT a correct explanation for statement -1 (c) Statement - 1 is True , Statement -2 is False (d) Statement -1 is False, Statement -2 is True |
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Answer» Correct Answer - B Statement -1 is true . For a moving observer, the near by objects appear to move in the opposite direction at a large speed . This is because the angular speed of the near by object w.r.t. observer is large . As the object moves away the angular velocity decreases and therefore its speed seems to be less . The distant object almost remains stationary. Statement -2 is the concept of relative velocity which states that ` vec v_(21) = vec v_(2G) - vecv_(1G) ` where `G` is the laboratory frame . Thus both the statement are true statement - 2 is not the correct explanation of statement -1. |
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| 174. |
A projectile is thrown upward with a velocity `v_(0)` at an angle `alpha` to the horizontal. The change in velocity of the projectile when it strikes the same horizontal plane isA. `v_(0) sin alpha` vertically downwardB. `2v_(0) sin alpha` vertically downwardC. `2v_(0) sin alpha` vertically upwardD. zero |
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Answer» Correct Answer - B `Deltav = a Deltat` (as a = constant) `= (-g hatj) ((2v_(0) sin alpha)/(g)) = (-2v_(0) sin alpha)hatj` i.e., change in velocity is `2v_(0) sin alpha`, vertically downwards. |
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| 175. |
A particle moves in the XY-plane according to the law `x = kt, y = kt (1- alphat)`, where k and `alpha` are positive constants and t is time. The trajectory of the particle isA. `y = kx`B. `y = x - (alpha x^(2))/(k)`C. `y =- (ax^(2))/(k)`D. `y = alpha x` |
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Answer» Correct Answer - B `x = kt, t = (x)/(k)` Now, `y = k ((x)/(k)) (1-alpha.(x)/(k))` or `y = x - (alpha x^(2))/(k)` |
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| 176. |
The height `y` and distance `x` along the horizontal plane of a projectile on a certain planet are given by `x = 6t m` and `y = (8t^(2) - 5t^(2))m`. The velocity with which the projectile is projected isA. `8 ms^(-1)`B. `9 ms^(-1)`C. `10 ms^(-1)`D. `(10//3) ms^(-1)` |
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Answer» Correct Answer - C `v_(y) = (dy)/(dt) = 8 - 10t, v_(x) = (dx)/(dt) = 6` At `t = 0, v_(y) = 8 ms^(-1)` and `v_(x) = 6 ms^(-1)` `:. V = sqrt(v_(x)^(2) +v_(y)^(2)) = 10 ms^(-1)` |
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| 177. |
The ratio of the speed of a projectile at the point of projection to the speed at the top of its trajectory is x. The angle of projection with the horizontal isA. `sin^(1) x`B. `cos^(-1)x`C. `sin^(-1) (1//x)`D. `cos^(-1) (1//x)` |
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Answer» Correct Answer - D Speed of projectile at the point of projection = u Speed of projectile at the top of its trajectory `= u cos theta`. `(u)/(u cos theta) = x` or `cos theta = (1)/(x) rArr theta = cos^(-1)((1)/(x))` |
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| 178. |
The equation of trajectory of an oblique projectile `y = sqrt(3) x - (g x^(2))/(2)`. The angle of projection isA. `90^(@)`B. zeroC. `60^(@)`D. `30^(@)` |
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Answer» Correct Answer - C Compare the given equation with `y = x tan theta - (gx^(2))/(2u^(2) cos^(2) theta) rArr tan theta = sqrt(3) rArr theta = 60^(@)` |
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| 179. |
An object is projected with a velocity of `20(m)/(s)` making an angle of `45^@` with horizontal. The equation for the trajectory is `h=Ax-Bx^2` where h is height, x is horizontal distance, A and B are constants. The ratio A:B is (g`=ms^-2)`A. `1:5`B. `5:1`C. `1:40`D. `40:1` |
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Answer» Correct Answer - D Standard equation of projectile motion `y = x tan theta - (gx^(2))/(2u^(2) cos^(2)theta)` Comparing with given equation `A = tan theta` and `B = (g)/(2u^(2) cos^(2) theta)` So, `(A)/(B) = (tan theta xx 2u^(2) cos^(2) theta)/(g) = 40` (As, `theta = 45^(@), u = 20 ms^(-1), g = 10 ms^(-2))` |
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| 180. |
A car , moving with a speed of `50 km//hr` , can be stopped by brakes after at least ` 6m `. If the same car is moving at a speed of `100 km//hr`, the minimum stopping distance isA. `12 m `B. `18 m`C. `24 m`D. ` 6 m` |
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Answer» Correct Answer - C Case - 1 : ` u = 50 xx (5)/(18) m//s`, `v = 0 , s = 6m , a = a` `v^(2) - u^(2) = 2as rArr 0^(2) -(50 xx (5)/(18))^(2) = 2 xx a xx 6` `rArr -(50 xx (5)/(18))^(2)` = 2 xx axx 6` …(i) Case - 2: `u = 100xx5/18 m//sec, v = 0 , s = s, a =a ` `:. v^(2) -u^(2) = 2as ` `rArr 0^(2)-(100xx5/18)^(2) = 2as ` `rArr -(100xx5//18)^(2) = 2 as` ...(ii) Dividing (i) and (ii) we get ` (100xx100)/(50xx50) = (2xxaxxs)/(2xxaxx6) rArr s = 24 m ` |
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| 181. |
What is average speed? |
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Answer» Average speed is the distance covered in unit time. |
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| 182. |
What is displacement? |
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Answer» Displacement is the shortest distance covered by the object in a specified direction. Its unit in the SI system is ‘meter’. |
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| 183. |
A particle is moving such that its position coordinates `(x, y)` are `(2m, 3m)` at time `t=0, (6m, 7m)` at time `t=2 s`, and `(13 m, 14m)` at time `t=5 s`. Average velocity vector`(vec(V)_(av))` from `t=0` to `t=5 s` isA. `(1)/(5) (13 hati +14 hatj)`B. `(7)/(3)(hati +hatj)`C. `2(hati - hatj)`D. `(11)/(5)(hati +hatj)` |
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Answer» Correct Answer - D Average velocity, `v_(av) = ("Net displacement")/("Time taken")` `= ((13 -2)hati+(14-3)hatj)/(5)` `= (11 hati + 11 hatj)/(5) = (11)/(5) (hati +hatj)` |
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| 184. |
A boy throws a ball with a velocity u at an angle `theta` with the horizontal. At the same instant he starts running with uniform velocity to catch the ball before it hits the ground. To achieve this he should run with a velocity ofA. `u cos theta`B. `u sin theta`C. `u tan theta`D. `u sec theta` |
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Answer» Correct Answer - A Velocity of boy should be equal to the horizontal component of velocity of ball i.e., `u cos theta`. |
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| 185. |
A train `100 m` long moving on a straight level track passes a pole in `5 s`. Find (a) the speed of the train (b) the time it will take to cross a bridge `500 m` long. |
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Answer» Correct Answer - (a) `20 m//s` , (b) `30s` (a) velocity `= ("distance travelled")/("time taken")` , i.e ., `v = ( 100 m)/( 5 s) = 20 m//s` (b) length of bridge `= 500 m`, length of train `= 100 m` In crossing the bridge, distance moved, `s = 500 + 100 = 600 m` , time , `t = (s)/( v) = (600)/(20) = 30s` |
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| 186. |
Which motion is called uniform? Why? |
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Answer» The motion of the body is said to be in uniform when its velocity is constant. |
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| 187. |
Is the direction of motion constant? How? |
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Answer» No, the direction of motion also changes continuously. |
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| 188. |
Can you give few examples for motion of an object where its speed remains constant but velocity changes? |
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Answer» For the bodies which are in uniform in circular motion the speed remain constant but velocity change. Ex : Rotation of earth, revolution of moon around the earth, etc. |
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| 189. |
What is true about scalar quantities ?(A) Scalars quantities have direction also. (B) Scalars can be added arithmetically.(C) There are special law to add scalars. (D) Scalars have special method to represent. |
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Answer» (B) Scalars can be added arithmetically. |
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| 190. |
A body is said to be in motion if :(A) Its position with respect to surrounding objects remains same(B) Its position with respect to surrounding objects keep on changing(C) Both (A) and (B)(D) Neither (A) nor (B) |
| Answer» (B) Its position with respect to surrounding objects keep on changing | |
| 191. |
In case of a moving body :(A) Displacement > Distance (B) Displacement < Distance(C) Displacement ≥Distance (D) Displacement ≤ Distance |
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Answer» (D) Displacement ≤ Distance |
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| 192. |
A body whose position with respect to surrounding does not change, is said to be in a state of :(A) Rest (B) Motion (C) Vibration (D) Oscillation |
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Answer» Is said to be in a Rest |
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| 193. |
If for the same range, the two heights attined are 20 m and 80 m, then the range will beA. 20 mB. 40 mC. 120mD. 160 m |
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Answer» Correct Answer - D According to given condition, `(h_(1))/(h_(2)) = (sin^(2)theta)/(sin^(2)(90^(@)-theta)) = (20)/(80)` `rArr tan^(2)theta = (1)/(4) rArr tan theta = (1)/(2)` `sin theta = (1)/(sqrt(5))` and `cos theta = (2)/(sqrt(5))` So, `h = (u^(2)sin^(2)theta)/(2g)` `rArr 20 = (u^(2))/(10g)` or `(u^(2))/(g) = 200` `:.` The range, `R = (u^(2) xx 2 sin theta cos theta)/(g) = (200 xx 2 xx 2)/(sqrt(5) xx sqrt(5))` `= (200 xx 4)/(5) = 160 m` |
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| 194. |
Assertion At height 20 m from ground , velocity of a projectile is `v = (20 hati + 10 hatj) ms^(-1)`. Here, `hati` is horizontal and `hatj` is vertical. Then, the particle is at the same height after 4s. Reason Maximum height of particle from ground is 40m (take, `g = 10 ms^(-2))`A. If both Asseration and Reason are correct and Reason is the correct explanation of AssertionB. If both Assertion and Reason are correct but Reason is not the correct explanation of AssertionC. If Assertion is true but Reason is falseD. If Assertion is false but Reason is true |
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Answer» Correct Answer - B `t = (2u_(y))/(g) = (2xx 20)/(10) = 4s` `H = 20 +(u_(y)^(2))/(2g) = 20 +((20)^(2))/(2 xx 10) = 40m` |
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| 195. |
A particle starts from origin at `t = 0` with a velocity of `15 hati ms^(-1)` and moves in xy-plane under the action of a force which produces a constant acceleration of `15 hati + 20 hatj ms^(-2)`. Find the y-coordinate of the particle at the instant its x-coordinate is 180 m. |
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Answer» The position of the particle is given by `r(t) = v_(0)t +(1)/(2) a t^(2) = 15 hati t +(1)/(5) (15 hati + 20 hatj) t^(2)` `= (15 t + 7.5 t^(2)) hati + 10 hatj t^(2)` `:. X(t) = 15 t +7.5 t^(2)` and `y(t) = 10 t^(2)` If `x(t) = 180m,t = ?` `180 = 15 t +7.5 t^(2) rArr t = 4s` `:.` y-coordinate, `y(t) = 10 xx 16 = 160 m` |
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| 196. |
The velocity of a projectile when it is at the greatest height is `(sqrt (2//5))` times its velocity when it is at half of its greatest height. Determine its angle of projection.A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `37^(@)` |
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Answer» Correct Answer - C `(u cos alpha) = sqrt((2)/(5)) sqrt((u cos alpha^(2))+{(u sin alpha)^(2)-2gh))}` Here, `h = (H)/(2) = (u^(2)sin^(2)alpha)/(4g)` Solving this equation we get,`alpha = 60^(@)` |
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| 197. |
A particle is thrown with a speed u at an angle ` theta` with the horizontal. When the particle makes an angle `phi` with the horizontal. Its speed changes to v :A. `v = u cos theta`B. `v = u cos theta cos phi`C. `v = u cos theta sec phi`D. `v = u sec theta cos phi` |
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Answer» Correct Answer - C Horizontal component of velocity remains uncanged. Hence, `v cos phi = u cos theta` `:. v = u cos theta sec phi` |
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| 198. |
A particle is projected in `x-y` plane with `y-`axis along vertical, the point of projection being origin. The equation of projectile is `y = sqrt(3) x - (gx^(2))/(2)`. The angle of projectile is ……………..and initial velocity is ………………… .A. `1ms^(-1)`B. `2ms^(-1)`C. `3ms^(-1)`D. `1.2 ms^(-1)` |
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Answer» Correct Answer - B Comparing with `y = x tan theta - (gx^(2))/(2u^(2)cos^(2)theta)` we have, `tan theta = sqrt(3)` or `theta = 60^(@)` and `u^(2) cos^(2)theta = 1` or `u = sec theta = sec 60^(@) = 2 ms^(-1)` |
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| 199. |
A boy goes from his house to school by bus at a speed of 20 km / h and returns back through the same route at a speed of 30 km / h The average speed of his journey is1. 24 km / h2. 25 km / h3. 30 km / h4. 20 km / h |
| Answer» 1. 24 km / h | |
| 200. |
The earth attracts a body of mass 1 kg with a force of 10 N. The mass of a boy is 50 kg. His weight will be 1. 50 kg 2. 500 N 3. 50 N 4. 5 N |
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Answer» His weight will be |
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