Tag: substances, objects and energy

Questions Related to substances, objects and energy

A ball of mass 50 g is thrown upwards. It rises to a maximum height of 100 m. At what height its kinetic energy will be reduced to 70%?

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  1. 30 m

  2. 40 m

  3. 60 m

  4. 70 m

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Correct Option: A
Explanation:

By work energy theorem,

$U _i+K _i=U _f+K _f$
$0+K _i=mgh+0$
$K _i=(50\times10^{-3})(10)(100)J$
$K _i=50J$ $-(i$)
Now, we have $K _f=\dfrac{70}{100}K _i$
By work energy theorem,
$0+K _i=mgh+\dfrac{70}{100}K _i$
$K _i-\dfrac{70}{100}K _i=(50\times10^{-3})(10)(h)$
$\dfrac{30}{100}K _i=0.5h$
From $(i)$, we have $K _i=50$
$h=2\times\dfrac{30}{100}\times50m$
$h=30m$

The kinetic energy K of a particle moving along a circle of radius R depends on the distance s as $ K=as^2$. The force acting on the particle is 

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  1. $\dfrac{2 as^2}{R}$

  2. $2a[1+\dfrac{s^2}{R^2}]^{1/2}$

  3. $2as$

  4. $2a[\dfrac{R^2}{s}]^{1/2}$

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Correct Option: A

The total energy of an electron is 3.555 MeV then its kinetic energy is

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  1. $3.545 \mathrm { MeV }$

  2. $3.045 \mathrm { MeV }$

  3. $3.5 \mathrm { MeV }$

  4. none

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Correct Option: B
Explanation:

Every substance has energy due to its mass$:$

loss in energy
$\Delta E = \Delta m{c^2}$
$\therefore KE = \left( {m - {m _0}} \right){c^2}$
$ = 3.555 - {m _0}{c^2}$
$ = 3.555 - 0.51$
$ = 3.045MeV$
Hence,
option $(B)$ is correct answer.

A steady beam of alpha-particles $(p=2e)$ traveling with constant kinetic energy $20$ $MeV$ carries a current $0.251$ x $10^{-6}$. If the beam is directed perpendicular to a plane surface, how many alpha-particles strike the surface in $3$ secs? 

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  1. $2.3$ x $10^{12}$

  2. $2.3$ x $10^{19}$

  3. $2.3$ x $10^{8}$

  4. $2.3$ x $10^{6}$

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Correct Option: C

A proton and a deutron both enter in a region uniform magnetic field B, moving at right angles to the field $\vec { \mathrm { B } }$  . If the radius of circular orbits for both the particles is equal and kinetic energy acquired by deutron is I Me V, then kinetic energy acquired by the proton particle will be . 

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  1. 1 MeV

  2. 2 MeV

  3. 0.5 MeV

  4. 4 MeV

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Correct Option: A
Explanation:

$R = \dfrac{{mv}}{{qB}} = \dfrac{{\sqrt {2mk} }}{{qB}}{K _p} = \dfrac{{{q^2}{B^2}R _p^2}}{{2mp}}$

$k = \dfrac{{{q^2}{B^2}{R^2}}}{{2m}}$
$\therefore \dfrac{{{K _\infty }}}{{{K _p}}} = {\left( {\dfrac{{{q _\infty }}}{{q{I _0}}}} \right)^2}\left( {\dfrac{{{m _p}}}{{{m _\infty }}}} \right){\left( {\dfrac{{{R _p}}}{{{R _\infty }}}} \right)^2}$
$ \Rightarrow {K _\infty } = {K _p}{\left( {\dfrac{{{q _\infty }}}{{{q _p}}}} \right)^2}\left( {\dfrac{{mp}}{{{m _\infty }}}} \right){\left( {\dfrac{{{R _\infty }}}{{{R _p}}}} \right)^2}$
${K _p} = 1MeV{\left( 2 \right)^2}{\left( {\dfrac{1}{4}} \right)^2}{\left( 1 \right)^2} = 1MeV$
Hence,
option $(A)$ is correct answer.

If K.E. body is increased by 100 % then % change in its momentum is:

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  1. 50%

  2. 41.4%

  3. 10%

  4. 20%

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Correct Option: A

A 200 gm mass has velocity of $(3\hat i+4\hat j)$m/s at certain instant. Find its kinetic energy. 

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  1. $2.5 J$

  2. $0.5 J$

  3. $0.8 J$

  4. $1.5 J$

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Correct Option: A
Explanation:

First we will calculate the Magnitude of velocity  so $v=\sqrt{3^2+4^2}=5\ m/s$

Now m=200 gm or 0.2 kg

So $KE=\dfrac{mv^2}{2}=\dfrac{0.2\times 5^2}{2}$

so $KE=2.5\ J$

A proton is kept at rest. A positively charged particle is released from rest at a distance $d$ in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time $t$, the work done on the two moving charged particles is:

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  1. same as the same force law is involved in the two experiments

  2. less for the case of a positron, as the positron moves away more rapidly and the force on it weakens

  3. more for the case of a positron, as the positron moves away a larger distance.

  4. same as the work done by charged particle on the stationary proton

Show answer
Correct Option: C
Explanation:
 The force between two protons is equal to the force between the proton and a position because their charges are same. As the mass of positron is much lesser than a proton, (1/1840 times) it moves away through much larger distance compared to proton. Change in their momentum will be same. So, a velocity of a lighter particle will be greater than that of a heavier particle. So, a positron is moved through a larger distance.
As $work done = force \times distance$. As forces are same in case of proton and positron but distance moved by positron is larger, hence, work done will be more.

An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because

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  1. The two magnetic forces are equal and opposite, so they produce no net effect

  2. The magnetic forces do no work on each particle

  3. The magnetic forces do equal and opposite (but non-zero) work on each particle

  4. The magnetic forces are necessarily negligible

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Correct Option: B
Explanation:

As the magnetic forces due to motion of electron and proton act in a direction perpendicular to the direction of motion, no work is done by these forces. That is why one ignores the magnetic force of one particle on another.

A car starts from rest and moves on a surface in which the coefficient of friction between the road and the tyres increases linearly with distance (x). If the car moves with the maximum possible acceleration, the Kinetic energy (E) of the car will depend on x as:

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  1. $E\propto\dfrac{1}{x^2}$

  2. $E\propto\dfrac{1}{x}$

  3. $E\propto{x}$

  4. $E\propto{x^2}$

Show answer
Correct Option: D
Explanation:

Support car moves a distance dx at a distance x.

$\mu=kx$
$f _r=kx(mg)$
Work done by friction $=K.E(E)$
$E=f _r dx=mg kx dx$
$E \propto x^2$