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A player caught a cricket ball of mass 150 g  moving  at a rate of 20 m/s . If the catching process is completed in 0.1 s  the force of the blow exerted by the ball on the hand of the player is equal to

  • Option 1)

    300 N

  • Option 2)

    150 N

  • Option 3)

    3 N

  • Option 4)

    30 N

 

30 N

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Posted by

Aadesh

A uniform cylinder of length L and mass M having cross - sectional area A is suspended, with its length vertical,from a fixed point by a massless spring,such that it is half submerged in a liquid of density \sigma at equilibrium position. The extension x_{0} of the spring when it is in equilibrium is :

 

 

 

  • Option 1)

    \frac{Mg}{k}\left ( 1+\frac{LA\sigma }{M} \right )

    (Here k is spring constant)

  • Option 2)

    \frac{Mg}{k}

    (Here k is spring constant)

  • Option 3)

    \frac{Mg}{k}\left ( 1-\frac{LA\sigma }{M} \right )

    (Here k is spring constant)

  • Option 4)

    \frac{Mg}{k}\left ( 1-\frac{LA\sigma }{2M} \right )

    (Here k is spring constant)

 

As we have learned @2013

At equilibrium F_{net } = 0

\Rightarrow Mg = k x_0 + \sigma \left ( \frac{AL}{2} \right )g

 

\Rightarrow kx_0 = Mg - \frac{\sigma AL}{2M}


Option 1)

\frac{Mg}{k}\left ( 1+\frac{LA\sigma }{M} \right )

(Here k is spring constant)

Option 2)

\frac{Mg}{k}

(Here k is spring constant)

Option 3)

\frac{Mg}{k}\left ( 1-\frac{LA\sigma }{M} \right )

(Here k is spring constant)

Option 4)

\frac{Mg}{k}\left ( 1-\frac{LA\sigma }{2M} \right )

(Here k is spring constant)

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Posted by

SudhirSol

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A heavy box is to be dragged along a rough horizontal floor. To do so, person A pushes it at an angle 300 from the horizontal and requires a minimum force FA, while person B pulls the box at an angle 600 from the horizontal and needs minimum force FB.If the coefficient of friction between the box and the floor is  \frac{\sqrt{3}}{5}  the ratio   \frac{F_{A}}{F_{B}}    is       

  • Option 1)

    \sqrt{3}

  • Option 2)

    \frac{5}{\sqrt{3}}

  • Option 3)

    \sqrt{\frac{3}{2}}

  • Option 4)

    \frac{2}{\sqrt{3}}

 

As learnt (Concept missing)

Given 1: person pushes box

Fa cos 30= (mg+Fa sin 30)M

Fa=Mmg \left | (\cos 30-\sin 30M) -------------- (1)

 

Case 2. Person pulls box

Fb \: \cos 60=Mmg \left | (\cos 60+M \sin 60) ------------------- (2)

\frac{(1)}{(2)}

\frac{Fa}{Fb}=\frac{2}{\sqrt{3}}


Option 1)

\sqrt{3}

This option is incorrect

Option 2)

\frac{5}{\sqrt{3}}

This option is incorrect

Option 3)

\sqrt{\frac{3}{2}}

This option is incorrect

Option 4)

\frac{2}{\sqrt{3}}

This option is correct

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Posted by

SudhirSol

Three masses m, 2m and 3m are moving in x-y plane with speed 3u, 2u, and u respectively as shown in figure. The three masses collide at the same point at P and stick together. The velocity of resulting mass will be :

  • Option 1)

    \frac{u}{12}(\hat{i}+\sqrt{3}\hat{j})

  • Option 2)

    \frac{u}{12}(\hat{i}-\sqrt{3}\hat{j})

  • Option 3)

    \frac{u}{12}(-\hat{i}+\sqrt{3}\hat{j})

  • Option 4)

    \frac{u}{12}(-\hat{i}-\sqrt{3}\hat{j})

 

As we learnt in 

Law of Consevation of Momentum -

 \vec{F}=\frac{\vec{dp}}{dt}

\vec{F}=0            then \vec{p}=constant

\vec{p}=\vec{p}_{1}+\vec{p}_{2}+\cdots =const

- wherein

\ast Independent of frame of reference

 

 

Pi = Py

m3u\hat{i}+3m[-ucos60\hat{i}+usin60\hat{j}]+2m[-2ucos60\hat{i}-2usin60\hat{j}]

=3mu\hat{i}+3m[\frac{-v}{2}\hat{i}+\frac{0\sqrt{3}\hat{j}}{2}]\Rightarrow \frac{u}{12}(\hat{-i}-\sqrt{3}\hat{j})


Option 1)

\frac{u}{12}(\hat{i}+\sqrt{3}\hat{j})

Incorrect

Option 2)

\frac{u}{12}(\hat{i}-\sqrt{3}\hat{j})

Incorrect

Option 3)

\frac{u}{12}(-\hat{i}+\sqrt{3}\hat{j})

Incorrect

Option 4)

\frac{u}{12}(-\hat{i}-\sqrt{3}\hat{j})

Correct

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Posted by

prateek

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 A small ball of mass m starts at a point A . with speed vo and moves along a frictionless track AB as shown. The track BC has coefficient of friction \mu. The ball comes to stop at C after travelling a distance L which is :

  • Option 1)

    \frac{2h}{\mu }+\frac{v_{0}^{2}}{2\mu g}

  • Option 2)

    \frac{h}{\mu }+\frac{v_{0}^{2}}{2\mu g}

  • Option 3)

    \frac{h}{2\mu }+\frac{v_{0}^{2}}{\mu g}

  • Option 4)

    \frac{h}{2\mu }+\frac{v_{0}^{2}}{2\mu g}

 

As we learnt in

Impulse Momentum Theorem -

\vec{F}=\frac{d\vec{p}}{dt}

\int_{t_{1}}^{t_{2}}\vec{F}dt=\int_{p_{1}}^{p_{2}}\vec{dp}

- wherein

If \bigtriangleup t  is increased, average force is decreased

 

 Apply convervation of Energy

mgh+\frac{1}{2}mV_{o}^{2}=\frac{1}{2}mV^{2}

V^{2}=Vo^{2}+2gh

Let a be the acceleration produced because of friction

V^{2}=u^{2}+2as=>0=Vo^{2}+2gh-2al

L=\frac{Vo^{2}+2gh}{2a}

f=MN  =>  Mmg

a = Mg

L=\frac{Vo^{2+2gh}}{2Mg}=\frac{h}{M}+\frac{Vo^{2}}{2Mg}

 

 


Option 1)

\frac{2h}{\mu }+\frac{v_{0}^{2}}{2\mu g}

This is incorrect option

Option 2)

\frac{h}{\mu }+\frac{v_{0}^{2}}{2\mu g}

This is incorrect option

Option 3)

\frac{h}{2\mu }+\frac{v_{0}^{2}}{\mu g}

This is correct option

Option 4)

\frac{h}{2\mu }+\frac{v_{0}^{2}}{2\mu g}

This is incorrect option

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Posted by

Aadil

When a constant force is applied to a body, it moves with uniform:

  • Option 1)

    Acceleration

  • Option 2)

    Velocity

  • Option 3)

    Speed         

  • Option 4)

    Momentum

 

As we discussed in

Newton's 2nd Law -

F\propto \frac{dp}{dt}

F=\tfrac{kdp}{dt} 

F=\tfrac{d\left (mv \right )}{dt} 

F=\tfrac{m\left (dv \right )}{dt}

\frac{dv}{dt}=a

Therefore  F=ma

- wherein

K=1 in C.G.S & S.I

 

 when a constant force is applied on a body, the body's accelerated with uniform acceleration.


Option 1)

Acceleration

Correct

Option 2)

Velocity

Incorrect

Option 3)

Speed         

Incorrect

Option 4)

Momentum

Incorrect

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Posted by

Plabita

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What is the mechanical advantage of single fixed pulley?

  • Option 1)

    1

  • Option 2)

    2

  • Option 3)

    0.5

  • Option 4)

    4

 

As we discussed in

Motion of connected blocks over pulley -

Equation of motion for   m_{1}

F_{net}=T-m_{1}g=m_{1}a

Equation of Motion for   m_{2}

F_{net}=m_{2}g-T=m_{2}a

 

- wherein

a=\frac{[m_{2}-m_{1}\:]g}{m_{1}+m_{2}}

T=\frac{2m_{1}m_{2}\:g}{m_{1}+m_{2}}


 

 

 


Option 1)

1

Correct

Option 2)

2

Incorrect

Option 3)

0.5

Incorrect

Option 4)

4

Incorrect

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Posted by

Plabita

Two bodies of masses 4kg and 5kg are acted upon by the same force. If the acceleration of lighter body is 2ms^{-2}, acceleration of heavier body is:

  • Option 1)

    4.2 ms^{-2}

  • Option 2)

    3.6 ms^{-2}

  • Option 3)

    2.4 \, ms^{-2}

  • Option 4)

    1.6\, ms^{-2}

 

As we discussed in

When 2 Blocks are in Contact -



F-f=m_{1}a

f=m_{2}a

- wherein

a=\frac{F}{m_{1}+m_{2}}

f=\frac{m_{2}F}{m_{1}+m_{2}}

 

 F=m_1a_1=4\times 3=12N

a_2=\frac{F}{m_2}=\frac{12}{5}=2.4 ms^{-2}

when same force applied on both bodies tried together

a_2=\frac{F}{m_1+m_2}=\frac{12}{4+5}=\frac{12}{9}=1.33 ms^{-2}


Option 1)

4.2 ms^{-2}

Incorrect

Option 2)

3.6 ms^{-2}

Incorrect

Option 3)

2.4 \, ms^{-2}

Incorrect

Option 4)

1.6\, ms^{-2}

Correct

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Posted by

Vakul

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Two blocks of masses 5kg and 10kg are connected to a pulley as shown in fig. What will be the acceleration if the pulley is set free?

  • Option 1)

    g

  • Option 2)

    g/2

  • Option 3)

    g/3

  • Option 4)

    g/4

 

(M1-M2)G/M1+M2.     =10-5×g/10+5=5g/15=g/3

 

Option 3

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Posted by

K nikil

Three blocks are connected as shown on a horizontal frictionless surface , if m_1=1kg \: \: , m_2= 8kg\: \: , m_3= 27kg\: \: and\: \: T_3=36N, \: \: T_2 will be?

  • Option 1)

    18N

  • Option 2)

    9N

  • Option 3)

    3.375N

  • Option 4)

    1.75 N

 

Answer 2

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Posted by

Siddharth Kapoor

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