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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.
| 101. |
On a horizontal curved road , the skidding or overturning of a vehicle will occur whenA. `v gt sqrt rg`B. the radius r is smallC. coefficient of friction `mu` is smallD. all of these |
| Answer» Correct Answer - D | |
| 102. |
For a body moving in a circular path, a condition for no skidding if `mu` is the coefficient of friction, isA. `(m^(2))/(r) ge mu mg`B. `(mv^(2))/(r) le mu mg`C. `(mv^(2))/(r)=mu mg`D. `v=r mu g` |
| Answer» Correct Answer - B | |
| 103. |
A traffic policeman standing at the intersection sees 2 cars A & B turning at angles `53^(@)&90^(@)` respectively (as shown in the figure). Their velocities are `V_(A)=20m//s`, `V_(B)=10m//s`. Then, which car appears to be moving faster to the traffic policeman:- A. AB. BC. Both equally fastD. Insufficient |
| Answer» Correct Answer - B | |
| 104. |
A body moves along a circular path of radius `5m`. The coefficient of friction between the surface of the path and the body is `0.5`. The angular velocity in `rad//s` with which the body should move so that it does not leave the path is `(g-10m//s^(-2))`A. 4B. 3C. 2D. 1 |
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Answer» Correct Answer - D `F=mu_(S)N,(mv^(2))/r=mumg` |
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| 105. |
In the above question, the magnitude of the average acceleration over the arc of `60^(@)` isA. `10 m//s^(2)`B. `11 m//s^(2)`C. `11.5 m//s^(2)`D. `12 m//s^(2)` |
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Answer» Correct Answer - C `theta=omegat` `t=(theta)/(omega)=(pi)/(3xx(v)/(r))=(pixxr)/(3v)=(pixx300)/(3xx60)` `t=(5pi)/(3)` `a_(av)=(Deltav)/(t)=(60)/((5pi)/(3))` `a_(av)=(60xx3)/(5pi)=(12xx3)/(pi)=11.5m//s^(2)`. |
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| 106. |
Assertion: A particle is rorating in a circle with constant speed as shown. Between point `A` and `B` , ratio of average acceleration and average velocity is angular velocity of particle about point `O` . lReason: Since speed is constant, angular velocity is also constant. A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, true but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - B `A` to `B` , `S=sqrt(2)R` and `|Deltav|=sqrt(v^(2)+v^(2)-2v.vcos90^(@))` `sqrt(2)v` Average acceleration `=(|Deltav|)/(t)=(sqrt(2)v)/(t)` And average velocity `=(S)/(t)=(sqrt(2)R)/(t)` The desired ratio is `(v)/(R)=omega` |
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| 107. |
In the figure, the block on the smooth table is set into motion in a circular orbit of radius r around the centre hole. The hanging mass is identicalm to the mass on the table and remains in equilibrium. Neglect friction. The string connecting the two blocks is massless and inextensible. Select the correct alternatives. A. The angular speed `omega` of the block in its circular motion is `sqrt((g)/(r))`B. Kinetic energy as function of r is given by `(mgr)/(2)`C. Angular momentum about the hole is conservedD. The block on table is in equilibrium |
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Answer» Correct Answer - A::B::C |
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| 108. |
Two bodies of masses m and 4 m are attached with string as shown in the figure. The body of mass `m` hanging from a string of length `l` is executing oscillations of angular amplitude `theta_(0)` while the other body is at rest. The minimum coefficient of friction between the mass `4m` and the horizontal surface should be A. `((2-cos theta_(0))/(3))`B. `2cos^(2)((theta_(0))/(2))`C. `((1-cos theta_(0))/(2))`D. `((3-2cos theta_(0))/(4))` |
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Answer» Correct Answer - D |
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| 109. |
Two bodies of masses `m` and `4m` are attached to a light string as shown in figure. A body of mass m hanging from string is executing oscillations with angular amplitude `60^@`, while other body is at rest on a horizontal surface. The minimum coefficient of friction between mass `4m` and the horizontal surface is (here pulley is light and smooth) A. `(1)/(4)`B. `(1)/(3)`C. `(1)/(2)`D. `(2)/(3)` |
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Answer» Correct Answer - C `h=l(1-cos 60^(@))=(1)/(2),v^(2)=2gl=gl` Now, `T_(max)-mg=(mv^(2))/(l)("at bottommost point")` `therefore" "T_(max)=2mg=mu_(s)(4mg)` `therefore" "mu_(s)=0.5` |
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| 110. |
A mass m on a friction less table is attached to a hanging mass M by a cord through a hole in the table . Then the angular speed with which m must spin for M stay at rest will be A. ` sqrt((Mg)/(mr))`B. `sqrt((mg)/(Mr))`C. `sqrt((mr)/(Mg))`D. `sqrt((g)/(r))` |
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Answer» Correct Answer - A C.F.=weight of body of mass M `mr omega^(2)=Mg` ` omega ^(2)=(Mg)/(mr)` `omega =sqrt((Mg)/(mr))` |
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| 111. |
A flywheel is revolving at 150 revolutions per minutes .If deccelerates at a constant rate of `2 pi rad //s^(2)`, then time requred to stop it isA. `10 s`B. `5 s`C. `2.5 s`D. `1.25 s` |
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Answer» Correct Answer - C `alpha =(omega_(2)-omega_(1))/(t)" "therefore t=(5pi)/(2pi)=2.5sec`. |
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| 112. |
If a bodyof mass 1000 gm is tied to free end of string of length 100 cm and whirled in a horizontal circle in a second, then the tecsion in the string will beA. `4pi^(2) N`B. `2pi^(2) N`C. `3pi^(2)N`D. `pi^(2)N` |
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Answer» Correct Answer - A `T=c.p.f.=m r omega ^(2)=m r ((2pi)/(T))^(2)` |
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| 113. |
A mass of 100 gm is tied to one end of a string 2 m long. The body is revolving in a horizontal circle making a maximum of 200 revolutions per min. The other end of the string is fixed at the centre of the circle of revolution. The maximum tension that the string can bear is (approximately)A. `8.76` NB. `8.94` NC. `89.42` ND. `87.64` N |
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Answer» Correct Answer - D Maximum tension `=momega^(2)r=mxx4pi^(2)xxn^(2)xxr` By substituting the values, we get `T_("max")=87.64N` |
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| 114. |
A vehicle is moving with uniform speed along horizontal, concave and convex surface roads. The surface on which, the normal reaction on the vehicle is maximum isA. horizontalB. concaveC. convexD. same on all surfaces |
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Answer» Correct Answer - B On horizontal surface, `R=mg` On concave shaped surface ,`R=mg+(mg^(2))/(r)` On convex shaped surface, `R=mg-(mg^(2))/(r)` |
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| 115. |
Two blocks of masses 1 kg and 2 kg are joined by a massless inextensible string of length 3m. Both blocks are kept on a horizontal table as shown. Friction coefficient between 2 kg block and table is zero. They are rotated about a vertical axis passing at a distance of 1m from 1 kg. Force of friction on 1kg block is (assume that there is enough friction between 1 kg block ground) A. 12 N towards centreB. 20 N towards centreC. 20 N away from the centreD. 12 N away from the centre |
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Answer» Correct Answer - D |
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| 116. |
A particle of mass m is being circulated on a vertical circle of radius r. If the speed of particle at the highest point be v, thenA. `mg=(mv^(2))/(r)`B. `mg gt(mv^(2))/(r)`C. `mgle(mv^(2))/(r)`D. `mg ge(mv^(2))/(r)` |
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Answer» Correct Answer - D If the speed of the particle at highest point be v, then `mg ge (mv^(2))/(r)` |
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| 117. |
A rotating wheel changes angular speed from 1800 rpm to 3000 rpm in 20 s. What is the angular acceleration assuming to be uniform?A. `60pi" rad s"^(-2)`B. `90pi" rad s"^(-2)`C. `2pi" rad s"^(-2)`D. `40pi" rad s"^(-2)` |
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Answer» Correct Answer - C We know that `omega=2pin" "therefore omega_(1)=2pin_(1)` where, `n_(1)=1800"rpm", n_(2)=3000"rpm", Deltat=20s` `rArr" "omega_(1)=2pixx(1800)/(60)=2pixx30=60pi` Similarly, `omega_(2)=2pin_(2)=2pixx(3000)/(60)=2pixx50=100pi` If the angular velocity of a rotating wheel about an axis changes from `omega_(1)` to `omega_(2)` in a time interval `Deltat`, then the angular acceleration of rotation wheel about that axis is given by `alpha=("change in angular velocity")/("time interval")` `=(omega_(2)-omega_(1))/(Deltat)=(100pi-60pi)/(20)=(40pi)/(20)=2pi" rad s"^(-2)` |
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| 118. |
A particle is moving along circular path of radius 40 m with a uniform speed of `20m//s`. Then the time taken for the particle to complete half revolution will beA. `4 pi s `B. `pi s`C. `2 pi s`D. `pi //2 s` |
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Answer» Correct Answer - C `omega = (v)/(r)=(20)/(40)=(1)/(2)` `T=(2pi)/(omega)=(2pi)/(1/2)=4pi` `T_(1//2)=(4pi)/(2)=2pi`. |
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| 119. |
A rotating wheel changes angular speed from 1800 rpm to 3000 rpm in 20 s. What is the angular acceleration assuming to be uniform?A. `60pi" rad s"^(-2)`B. `90pi" rad s"^(-2)`C. `2pi"rad s"^(-2)`D. `40pi"rad s"^(-2)` |
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Answer» Correct Answer - C We know that, `omega=2pir" "therefore_(1)=2pin_(1)` where, `n_(1)=1800 "rpm", n_(2)=3000"rpm", Deltat=20s` `omega_(1)=2pixx(1800)/(60)=2pixx30=60pi` Similarly , `omega_(2)=2pin_(2)=2pixx(3000)/(60)=2pixx50=100pi` If the angular velocity of a rotating wheel about an axis is changed by change in angular velocity in a time interval `Deltat`, then the angular acceleration of rotating wheel about that axis is given by `alpha=("Change in angular velocity")/("Time interval")` `=(omega_(2)-omega_(1))/(Deltat)=(100pi-60pi)/(20)` `=(40pi)/(20)=2pi"rad s"^(-2)` |
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| 120. |
In 20 seconds, the speed of a motor changes from 1200 rpm to 1800 .In this period , of number of revoutions completed by it isA. 500B. 400C. 200D. 100 |
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Answer» Correct Answer - A `N=(n_(1)+n_(2))/(2)xxt=25xx20=500`. |
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| 121. |
A ball of mass `500g` tied to one end of string is revolved in a horizontal circle of radius `10cm` with a speed `1//pirev//s` in gravity-free space, then the linear velocity, acceleration and tension in the string will beA. `0.1m//s,0.4m//s^(2),0.2N`B. `0.1m//s,04m//s^(2),0.1N`C. `0.2m//s,0.4m//s^(2),0.2N`D. `0.2m//s,0.3m//s^(2),0.2N` |
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Answer» Correct Answer - C `omega=2piN=2pixx(1)/(pi)=2rad//s` `m=0.5kg,r=0.1m` `v=romega=0.1xx2=0.2m//s` `a=omega^(2)r=(2)^(2)xx0.1=0.4m//s^(2)` `T=momega^(2)r=0.5xx0.4=0.2N` |
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| 122. |
The speed of a motor increases from 1200 rpm to 1800 rpm in 20 s . How many revolutions does it make during these second ?A. 400B. 600C. 500D. 700 |
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Answer» Correct Answer - C `theta =2pi((n_(1)+n_(2)))/(2)t` `thereforetheta =1000 pi N=(theta)/(2pi)` |
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| 123. |
A particle moves in a circle of radius 25 cm at two revolutions per sec. The acceleration of the particle in `m//s^(2)` is:A. `12 pi ^(2)`B. `8pi^(2)`C. `4pi^(2)`D. `2 pi^(2)` |
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Answer» Correct Answer - C `a=romega^(2)` `=r4pi^(2)n^(2)` `=0.25xx4pi^(2)xx4=4pi^(2)` |
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| 124. |
A string of lenth l fixed at one end carries a mass m at the other end. The strings makes `(2)/(pi)revs^(-1)` around the axis through the fixed end as shown in the figure, the tension in the string is A. 16 mlB. 4 mlC. 8 mlD. 2 ml |
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Answer» Correct Answer - A Balancing horizontal forces, `Tsintheta=mromega^(2)` or `Tsintheta=m(l sintheta)omega^(2)` `therefore" "T=mlomega^(2)=ml(2pif)^(2)` `=ml(2pixx(2)/(pi))^(2)=16ml` |
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| 125. |
If a body moves along circular path of constant radius, then the magnitude of its acceleration will beA. uniformB. zeroC. variableD. not to be decided from the information |
| Answer» Correct Answer - D | |
| 126. |
A particle moving on a circular path makes 600 rpm. In how much time it will complete one revolution?A. 0.2sB. 0.1sC. 0.4sD. 0.3s |
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Answer» Correct Answer - B Here, rpm mean rotation per minute, i.e., it is frequency of rotation `f="600 rpm"rArrf=(600)/(60)"rpm"=10Hz` `rArr" "T=(1)/(f)=(1)/(10)=0.1s` |
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| 127. |
If a particle moves on a circular path of radius 4 m with time period 4 s, then the change in magnitude of velocity in one- fourth revolution will beA. `2sqrt2 pi m//s`B. `pi m//s`C. `3 pi m//s`D. `4 pi m//s` |
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Answer» Correct Answer - A `Deltav=2v "sin" (theta)/(2)` `v=r omega =4 xx (2pi)/(4)=2pi` |
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| 128. |
A string of length `L` is fixed at one end and carries a mass `M` at the other end. The string makes `2//pi` revolution per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is. A. `ML`B. `2ML`C. `4ML`D. `16ML` |
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Answer» Correct Answer - D `omega=2piN=2pi((2)/(pi))=4rad//s` As in the previous solution, `T=Momega^(2)L=16ML` |
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| 129. |
If a car is travelling at 20 m//s on a circular road of radius 400m then the radial acceleration will beA. `1m//s^(2)`B. `10m//s^(2)`C. `0.1m//s^(2)`D. `0.01 m//s^(2)` |
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Answer» Correct Answer - A `a_(r)=(v^(2))/(2)=(20xx20)/(400)=1`. |
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| 130. |
A particle is moving in a circle of a radius 40 cm has a linear speed of `30m//s` at a certains intant its linear speed is increasesing at the rate of `4m//s^(2)`. Increasing at the instant will beA. `200m//s^(3)`B. `600m//s^(3)`C. `100m//s^(2)`D. `300m//s^(3)` |
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Answer» Correct Answer - B `(d)/(dt)(v^(2)/(v))=(2v)/(r)(dv)/(dt)` `=(2xx30xx4)/(0.4)=600m//s^(3)` |
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| 131. |
A particle moves in a circular path of radius 0.4 m with a constan speed . If it makes 5 revolution in each second of its motion , then the speed of the particle will beA. `10.6 m//s`B. `11.2 m//s`C. `12.56 m//s`D. `13.6 m//s` |
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Answer» Correct Answer - C `v= r omega ` |
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| 132. |
A motor car is travelling 20m/s on a circular road of radius 400 m. If it increases its speed at the rate of `1 m//s^(2)`, then its acceleration will beA. `2sqrt2m//s^(2)`B. `sqrt3m//s^(2)`C. `sqrt 2 m//s^(2)`D. `3 sqrt3 m//s^(2)` |
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Answer» Correct Answer - C `a_(r)=v^(2)(r)=(20xx20)/(400)=1," "a_(r)=1` `a_(R)=sqrt(a_(T)^(2)+a_(r)^(2))=sqrt(1+1)=sqrt2` |
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| 133. |
The angular speed of the minutes hand of a clock in degree per second isA. `0.01`B. `0.1`C. `0.0`D. `pi//1800` |
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Answer» Correct Answer - B `omega =(theta)/(T)` |
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| 134. |
If the speed of the tip of the minute hand of a town clock is `1.75xx10^(-3) m//s`, then the speed of its second hand of same length will beA. `1.7xx10^(-3) m//s`B. `10.510^(-3) m//s`C. `10.5xx10^(-2) m//s`D. `17.5xx10^(-3) m//s` |
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Answer» Correct Answer - C `v_(1) = r omega _(1) and v_(2)= r omega _(2) therefore (v_(2))/(v_(1))=(omega _(2))/(omega_(1))=(T_(1))/(T_(2))` |
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| 135. |
In a clock, what is the time period of meeting of the minute hand and the second hand ?A. `59s`B. `(60)/(59)s`C. `(59)/(60)s`D. `(3600)/(59)s` |
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Answer» Correct Answer - D `omega_(1)t-omega_(2)t=2pi` :. `t=(2pi)/(omega_(1)-omega_(2))` `=(2pi)/((2pi//T_(1))-(2pi//T_(2)))` `=(T_(1)T_(2))/(T_(2)-T_(1))` `=((3600)(60))/((3600)-(60))` `=(3600)/(59)s` |
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| 136. |
The hour hand and the minute hand of a clock coincide at every relative peridic time is ,A. `11//12 hour`B. `12//11 hour`C. `11//6 hour`D. `12//24 hour` |
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Answer» Correct Answer - B For relative motion, both the objects are moving in the same direction then their relative velocity is `omega_(rel)=omega_(2)-omega_(1)` `(2pi)/(T)=2pi [(1)/T_(2)-(1)/(T_(1))]=2pi[(T_(1)-T_(2))/(T_(1)T_(2))]` `therefore T=(T_(1) T_(2))/(T_(1)-T_(2))=(12xx10)/(12-1)=(12)/(11)h` |
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| 137. |
Calculate the average angular velocity of the hour hand of the of a clock.A. `(pi)/(43200) rad //s`B. `(pi)/(21600) rad//s`C. `(pi)/(30) rad//s`D. `(pi)/(1800) rad//s` |
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Answer» Correct Answer - B `omega =(2pi)/(T)=(2pi)/(12xx3600)=(pi)/(21600)` |
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| 138. |
A particle is moving in a circle of radius R with constant speed. The time period of the particle is T. In a time `t=(T)/(6)` Average velocity of the particle is…..A. `(3R)/(T)`B. `(6R)/(T)`C. `(2R)/(T)`D. `(4R)/(T)` |
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Answer» Correct Answer - B |
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| 139. |
A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force. (Consider `g=10m//s^2`).A. `20N`B. `220N`C. `4N`D. `16N` |
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Answer» Correct Answer - B `v^(2)-u^(2)=2as,F-mg=ma` |
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| 140. |
A track consists of two circular pars ABC andCDE of equal rdius 100 m and joined smoothly as shown in figure.Each part sutends a right ngle at its centre. A cycle weighing 100 kg together with rider travels at a constant speed of 18 km/h on the track. A. Findteh nromal contct force by tehroad on the cycle whenit is at B and at D. b.Findteh force of friction exerted by the track on the tyres when the cycle is at B,C and D. c. Find the normal force between teh road and teh cycle just before and just after the cycle crosses C. d. What should be the minimum friction coefficient between the road and the tyre, which will ensure that teh cyclist can move with constant speed? Take `g=10 m/s^2` |
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Answer» Correct Answer - A::B::C::D Radius of the curves =100 m ` Weight = 100 kg velocity = 18 km/hr=5m/sec a. At B, mg-(mv^2)/r=N` `rarrN=(100xx10)-(100xx25/100)` `=1000-25=975N` At D,` N=mg+(mv^2)/R` `=1000+25=1025N` b.At B & D, the cycle has no tendency to slide. So at B & D frictioN/Al is zero. At C `mg sintheta=f` `rarr 1000xx(1/sqrt2)=707N` `c.i Before C mg costheta-N=(mv^2)/r` `rarr N=mgcostheta-(mv^2)/r` =707-25=682N` ii. `N-mgcostheta=(mv^2)/r` `rarr N=(mv^2)/r+mgcostheta` `=25+707=732N` d. To find out the minimum desired coefficient of frictionwe have to consider a point just before c (where N is minimum). NOw `muN=mg sintheta` `rarr muxx682=707` So, `mu=1.037` |
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| 141. |
A small sphere `B` of mass `m` is released form rest in the position shown and swings freely in a vertical plane, first about `O` and then about the peg `A` after the cord comes in contact with the peg. Determime the tension in the cord (a) just before the sphere comes in contact with the peg. (b) just after it comes in contact with the peg. |
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Answer» Correct Answer - A::B::C::D `h=0.8sin30^(@)=0.4m` :. `v^(2)=2gh` (a) just before, `T_(1)-mgsin30^(@)=(mv^(2))/(R_(1))` , `(R_(1)=0.8m)` `T_(1)=(mg)/(2)+(m(2g)(0.4))/(0.8)` `=(3mg)/(2)` (b) just after, `T_(2)-mgsin30^(@)=(mv^(2))/(R_(2))` , `(R_(2)=0.4m)` `T_(2)=(mg)/(2)+(m(2g)(0.4))/(0.4)` or `T_(2)=(5mg)/(2)` |
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| 142. |
A smalll bob attached to a string of length `l` is suspended from a rigid support and rotates with uniform speed along a circle in a horizontal plane. Let `theta` be the angle made by the string with the vertical, then the length of a simple pendulum having the same period isA. `l//cos theta`B. `l sin theta`C. `l//sin theta`D. `l cos theta` |
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Answer» Correct Answer - D |
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| 143. |
Two different blocks of equal mass m are released from two positions as shown in figure Net force on the block at bottommost point in case (i) is say `F_(1)` and in case (ii) is say `F_(2)`. Then A. `F_(1)=F_(2)`B. `F_(1) gt F_(2)`C. `F_(1)ltF_(2)`D. Data insufficient |
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Answer» Correct Answer - C |
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| 144. |
A stone is rotated in a vertical circle. Speed at bottommost point is `sqrt(8gR)`, where R is the radius of circle. The ratio of tension at the top and the bottom isA. `1:2`B. `1:3`C. `2:3`D. `1:4` |
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Answer» Correct Answer - B `T=(m u^(2))/(R)+"mg cos"theta,v^(2)=u^(2)-2gh` `T_("top")=3mg" and "T_("bottom")=9mg` Therefore, the ratio `(T_("top"))/(T_("bottom"))=(3mg)/(9mg)=(1)/(3)` |
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| 145. |
In th efigure shown in fig. `10.33` , a bob attached with a light string of radius `R` is given as initial velocity `u=sqrt(4gR)` at the bottommost point. (a) At what height string will slack. (b) What is velocity of the bob just befare slacking of string. |
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Answer» Correct Answer - B::C String will slack at height `h_(1)` , discussed in article `10.5` where, `h_(1)=(u^(2)+gR)/(3g)=(5)/(3)R` , `(as u^(2)=4gR)` `v=sqrt(u^(2)-2gh_(1))=sqrt(4gR-2gxx(5)/(3)R)` `=sqrt((2)/(3)gR)` |
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| 146. |
Assertion: A pendulam is oscillating between points `A` , `B` and `C` . Acceleration of bob at points `A` or `C` is zero. Reason: Velocity at these points is zero.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, true but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - D At `A` and `C` , `v=0` . Therefore, radial acceleration `(v^(2))/(R)` is zero. But tangential acceleration `gsintheta` is non-zero. |
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| 147. |
Assertion A particle is roting in a circle of radius 1 m. At some given instant, its speed is `2 ms^(-1)`. Then acceleration of particle at the given instant is `4 ms^(-2)`. Reason Centripetal acceleration at this instant is `4 ms^(-2)` towards centre of circle.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.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. |
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Answer» Correct Answer - B `a_(c)=(v^(2))/(R)=((2)^(2))/(1)=4ms^(-2)` But no information is given for tangential acceleration `a_(t)`. |
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| 148. |
The velocity and acceleration vectors of a particle undergoing circular motion are `v = 2 hat(i) m//s and a = 2 hat(i)+4 hat(j)m//s^(2)` respectively at some instant of time. The radius of the circle isA. 1 mB. 2 mC. 3 mD. 4 m |
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Answer» Correct Answer - A |
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| 149. |
Assertion: Velocity and acceleration of a particle in circular motion at some instant are: `v=(2hat(i))ms^(-1)` and `a=(-hat(i)+2hat(j))ms^(-2)` , then radius of circle is `2m` . Reason: Speed of particle is decreasing at a rate of `1ms^(-2)` .A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, true but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - B Component of acceleration perpendicular to velocity is centripetal acceleration. :. `a_(c)=(v^(2))/(R)` or `R=(v^(2))/(a_(c))=((2)^(2))/(2)=2m` |
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| 150. |
At a curved path of a road, the road bed is raised the curved path, the slop of the road bed is given by the equation .A. ` tan theta =(v^(2))/(rg)`B. `tan theta (v^(2)r)/(g)`C. `tan theta =(rg)/(v^(2))`D. `tan theta =(g)/(vr^(2))` |
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Answer» Correct Answer - A `v^(2)= rg tan theta ` ` tan theta =(v^(2))/(rg)` |
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