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501.

A cyclist is riding with a speed of `27 km h^(-1)`. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate `0.5 ms^(-1)`. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?A. `0.86 m//s^(2)`B. `0.43 m//s^(2)`C. `1.24 m//s^(2)`D. `1.76 m//s^(2)`

Answer» Correct Answer - A
`v=27xx(5)/(18)=(45)/(6)=(15)/(2)`
` a=sqrt(a^(2)+((v^(2))/(r))^(2))=sqrt(0.25+(15xx15)/(4xx80))`
`=sqrt(0.25+(45)/(64))`
`=sqrt(0.25+0.49)=sqrt(0.25+0.70)=sqrt(0.95)`
`0.86m//s^(2)`
Discussion
502.

A car of mass 800 kg moves on a circular track of radius 40 m . If the coefficient of friction is 0.5, then maximum velocity with which the car can move isA. ` 7 m//s`B. `14 m//s`C. `8 m//s`D. `12m//s`

Answer» Correct Answer - B
`V=sqrt(mu rg)`
`=sqrt(0.5xx40xx9.8)`
`=sqrt(20xx9.6)=sqrt(196)=14 m//s.`
Discussion
503.

A cyslist is traveilling on a circular section of highway of radius 2500 ft at the speed of 60 mile/h. The cyclist suddenly applies the brakes causing the bicycle to slow down at constant rate. Knowing that after 8 second,the speed has been reduced to 45 mile/h. The acceleration of the bicyle immediately after the breakes have been applied. isA. `2 ft//s^(2)`B. `4.14 ft//s^(2)`C. `3.10 ft//s^(2)`D. `2.75 ft^(2)//s`

Answer» (b) Tangential acceleration
`=a_(t)=(v_(t)-v_(i))/(t)=((45-60)/(8))((22)/(15))=-(11)/(4)ft//s^(2)`
The radial acceleration is
`a_(n)=(v^(2))/(r)=(60xx(22)/(15))^(2)/(2500)=3.1`
`:." " a=sqrt(a_(n)^(2)+a_(t)^(2))=4.14 ft//s^(2)`
Discussion
504.

A car of 1400 kg kis moving on a circular path of radius 30 m with a speed of 40 km/h. When the drive applies the breaks and the carf conntinues to move along the circular path, what is the maximum deceleration possible if the types are limited to a horozontal friction of `10.6 kN`?A. `10 m//s^(2)`B. `6.36 m//s^(2)`C. `4 m//s^(2)`D. None

Answer» (b) the net acceleration to car is provided by force of friction.
`:. " " a=sqrt(a_(n)^(2)+a_(1)^(2))`
`:." "F=ma=msqrt(((v^(2))/(r))^(2)+a_(1)^(2))`
or `" " 10.6xx10^(3)=msqrt((v^(4))/(r^(2))+a_(t)^(2))`
After solving, " " `a_(t)=6..36 m//s^(2)`
Discussion
505.

A stationary shell breaks into three fragments The momentum of two of the fragments is `P` each and move at `60^(@)` to each other.The momentum of the third fragment isA. PB. 2PC. `P/sqrt3`D. `sqrt3P`

Answer» Correct Answer - D
`barP_(3)=-(barP_(1)+barP_(2)),p_(3)=sqrt(p_(1)^(2)+p_(2)^(2)+2p_(1)p_(2) cos theta)`
Discussion
506.

Find the accelerationof the moonwith respect to the eath from the following data, Distance between the earth and the moon `=3.85xx10^5` km and the time taken by the moon to complete one revolution aroudn the earth `=27.3days

Answer» Correct Answer - A::B::C
Distance between earth and Moon `
`r=3.85xx10^5km`
`=3.85xx10^8`m
`T=27.3 days `
`=24xx3600xx(27.3)sec`
`2.36xx10^6sec`
v=(2pir)/T` ltbr.gt `=(2xx3.14xx3.85xx10^8)/(2.36xx10^6)`
`=1025.42 m/sec `
`a=v^2/r= ((1025.420^2)/(2.36xx10^6)`
`=0.00273 m/sec^`
`=2.73xx10^3 m/sec^2`
Discussion
507.

A particle moves in circle of radius 1.0 cm at a speed given by v=2.0 t where v is in cm/s and t in seconeds. A. find the radia accelerationof the particle at t=1 s. b. Findthe tangential acceleration at t=1s. c.Find the magnitude of the aceleration at t=1s.

Answer» Correct Answer - A::B::C::D
`V=2,r=1cm`
a. Radial acceleration at t=1 sec
`a=(v^2/r)=((2)^2)/1=4cm/sec^2`
b. Tangential acceleration at t=1 sec`
a=(dv)/(dt)`
`=d/(dt) (2t) =2cm/sec^2`
c. Magnitude of acceleration at t=1sec
`a=sqrt(4^2+2^2)`
`=sqrt(20) cm/sec^2`
Discussion
508.

A marble block of mass 2 kg lying on ice when given a velocity of `6m//s` is stopped by friction in 10s. Then the coefficient of friction isA. `0.02`B. `0.03`C. `0.06`D. `0.01`

Answer» Correct Answer - C
`v=u+at,a=mu_(k)g`
Discussion
509.

A car of mass `1000kg` negotiates a banked curve of radius `90m` on a fictionless road. If the banking angle is `45^(@)` the speed of the car is:A. `20 ms^(-1)`B. `30 ms^(-1)`C. `5 ms^(-1)`D. `10 ms^(-1)`

Answer» Correct Answer - B
The angle of banking
`tantheta=(v^(2))/(rg)`
Given, `theta=45^(@), r=90n" and "g=10ms^(-2)`
`tan45^(@)=(v^(2))/(90xx10)rArrv=sqrt(90xx10xxtan45^(@))`
Speed of car, `v=30ms^(-1)`
Discussion
510.

A frictionless track ABCDE ends in a circular loop of radius R. A body slides down the track from point A which is at height h = 5cm. Maximum value of R for a body to complete the loop successfully is A. 2 cmB. `(10)/(3)`cmC. `(15)/(4)`cmD. `(18)/(3)`cm

Answer» Correct Answer - A
To complete the loop successfully `sqrt(2gh)=sqrt(5gR)`
`therefore" "R=(2h)/(5)=(2xx5)/(5)=2cm`
Discussion
511.

A particle of mass `m` attached to a string of length `l` is descending circular motion on a smooth plane inclined at an angle `alpha` with the horizontal. For the particle to reach the highest point its velocity at the lowest point should exceed.A. `sqrt(5gl)`B. `sqrt(5gl(cos alpha+1))`C. `sqrt(5gl//tan alpha)`D. `sqrt(5gl sin alpha)`

Answer» Correct Answer - D
Discussion
512.

An automobile enters a turn whose radius is R. The road is banked at angle `theta`. Fricton is negligible between wheels of the automobile and road, Mass of automobile is `m` and speed is `v`. Select the correct alternative.A. net force on the automobile is zeroB. normal reaction on the automobile is `mg cos theta`C. normal reaction on the automobile is `mg sec theta`D. net force on the automobile is `sqrt((mg)^(2)+(mv^(2)//R)^(2))`

Answer» Correct Answer - C
Discussion
513.

A ball of mass 1 kg is released from position A inside a wedge with a hemispherical cut of radius 0.5 m as shown in figure. The force exerted by the vertical wall OM on wedge, when the ball is in position B is (neglect friction everywhere) (Take `g = 10 m//s^(2)`) A. 10 NB. `5 sqrt(3) N`C. `(15 sqrt(3))/(2)N`D. 15 N

Answer» Correct Answer - C
Discussion