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PHXI04:MOTION IN A PLANE

361891 A wheel is rotating at $900 r p m$ about its axis. When the power is cut off, it comes to rest in $1 \min$ . The angular retardation (in $r a d s^{- 2}$ ) is

1

π / 2

2

π / 4

3

π / 6

4

π / 8

.

Explanation:

As, $ω = ω_{0} + α t$
Here, $ω_{0} = 900 rpm = \frac{(2 π \times 900)}{60} {rads}^{- 1}$ ,
$ω = 0$ and $t = 60 s$
So, $0 = \frac{2 π \times 900}{60} - α \times 60$
$0 = \frac{2 π \times 900}{60} - α \times 60$
$α = \frac{2 π \times 900}{60 \times 60} = \frac{π}{2}$

PHXI04:MOTION IN A PLANE

361892 A particle is moving in a radius $R$ with constant speed $v$ . The magnitude of average acceleration after half revolution is

1

\frac{2 π}{R v}

2

\frac{2 π}{R v^{2}}

3

\frac{2 v^{2}}{π R}

4

\frac{2 v}{π R^{2}}

Explanation:

After half revolution, change in velocity,
$Δ v = v - (- v) = 2 v$
Time taken for half revolution,
$Δ t = \frac{T}{2} = \frac{1}{2} \frac{2 π R}{v} = \frac{π R}{v}$
$∴$ Average acceleration after half revolution,
$a_{a v g} = \frac{Δ v}{Δ t} = \frac{2 v}{(\frac{π R}{v})} = \frac{2 v^{2}}{π R}$

MHTCET - 2020

PHXI04:MOTION IN A PLANE

361893 Two particles of equal masses are revolving in circular paths of radii $r_{1}$ and $r_{2}$ respectively with the same speed. The ratio of their centripetal forces is

1

\frac{r_{2}}{r_{1}}

2

\sqrt{\frac{r_{2}}{r_{1}}}

3

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

4

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

Explanation:

$F = \frac{m v^{2}}{r}$ If $m$ and $n$ are constants then
$F \propto \frac{1}{r}$
$∴ \frac{F_{1}}{F_{2}} = (\frac{r_{2}}{r_{1}})$

PHXI04:MOTION IN A PLANE

361894 If a particle covers half the circle of radius $R$ with constant speed $v$ then

1 Change in momentum is

m v r

2 Change in

K

.

E

. is

1 / 2 m v^{2}

3 Change in

K

.

E

. is

m v^{2}

4 Change in

K

.

E

. is zero

Explanation:

As momentum is vector quantity
$∴$ Change in momentum
$Δ P = 2 m v \sin (θ / 2)$
$= 2 m v \sin (90) = 2 m v$
But kinetic energy remains always constant so change in kinetic energy is zero.
supporting img

PHXI04:MOTION IN A PLANE

361891 A wheel is rotating at $900 r p m$ about its axis. When the power is cut off, it comes to rest in $1 \min$ . The angular retardation (in $r a d s^{- 2}$ ) is

1

π / 2

2

π / 4

3

π / 6

4

π / 8

.

Explanation:

As, $ω = ω_{0} + α t$
Here, $ω_{0} = 900 rpm = \frac{(2 π \times 900)}{60} {rads}^{- 1}$ ,
$ω = 0$ and $t = 60 s$
So, $0 = \frac{2 π \times 900}{60} - α \times 60$
$0 = \frac{2 π \times 900}{60} - α \times 60$
$α = \frac{2 π \times 900}{60 \times 60} = \frac{π}{2}$

PHXI04:MOTION IN A PLANE

361892 A particle is moving in a radius $R$ with constant speed $v$ . The magnitude of average acceleration after half revolution is

1

\frac{2 π}{R v}

2

\frac{2 π}{R v^{2}}

3

\frac{2 v^{2}}{π R}

4

\frac{2 v}{π R^{2}}

Explanation:

After half revolution, change in velocity,
$Δ v = v - (- v) = 2 v$
Time taken for half revolution,
$Δ t = \frac{T}{2} = \frac{1}{2} \frac{2 π R}{v} = \frac{π R}{v}$
$∴$ Average acceleration after half revolution,
$a_{a v g} = \frac{Δ v}{Δ t} = \frac{2 v}{(\frac{π R}{v})} = \frac{2 v^{2}}{π R}$

MHTCET - 2020

PHXI04:MOTION IN A PLANE

361893 Two particles of equal masses are revolving in circular paths of radii $r_{1}$ and $r_{2}$ respectively with the same speed. The ratio of their centripetal forces is

1

\frac{r_{2}}{r_{1}}

2

\sqrt{\frac{r_{2}}{r_{1}}}

3

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

4

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

Explanation:

$F = \frac{m v^{2}}{r}$ If $m$ and $n$ are constants then
$F \propto \frac{1}{r}$
$∴ \frac{F_{1}}{F_{2}} = (\frac{r_{2}}{r_{1}})$

PHXI04:MOTION IN A PLANE

361894 If a particle covers half the circle of radius $R$ with constant speed $v$ then

1 Change in momentum is

m v r

2 Change in

K

.

E

. is

1 / 2 m v^{2}

3 Change in

K

.

E

. is

m v^{2}

4 Change in

K

.

E

. is zero

Explanation:

As momentum is vector quantity
$∴$ Change in momentum
$Δ P = 2 m v \sin (θ / 2)$
$= 2 m v \sin (90) = 2 m v$
But kinetic energy remains always constant so change in kinetic energy is zero.
supporting img

PHXI04:MOTION IN A PLANE

361891 A wheel is rotating at $900 r p m$ about its axis. When the power is cut off, it comes to rest in $1 \min$ . The angular retardation (in $r a d s^{- 2}$ ) is

1

π / 2

2

π / 4

3

π / 6

4

π / 8

.

Explanation:

As, $ω = ω_{0} + α t$
Here, $ω_{0} = 900 rpm = \frac{(2 π \times 900)}{60} {rads}^{- 1}$ ,
$ω = 0$ and $t = 60 s$
So, $0 = \frac{2 π \times 900}{60} - α \times 60$
$0 = \frac{2 π \times 900}{60} - α \times 60$
$α = \frac{2 π \times 900}{60 \times 60} = \frac{π}{2}$

PHXI04:MOTION IN A PLANE

361892 A particle is moving in a radius $R$ with constant speed $v$ . The magnitude of average acceleration after half revolution is

1

\frac{2 π}{R v}

2

\frac{2 π}{R v^{2}}

3

\frac{2 v^{2}}{π R}

4

\frac{2 v}{π R^{2}}

Explanation:

After half revolution, change in velocity,
$Δ v = v - (- v) = 2 v$
Time taken for half revolution,
$Δ t = \frac{T}{2} = \frac{1}{2} \frac{2 π R}{v} = \frac{π R}{v}$
$∴$ Average acceleration after half revolution,
$a_{a v g} = \frac{Δ v}{Δ t} = \frac{2 v}{(\frac{π R}{v})} = \frac{2 v^{2}}{π R}$

MHTCET - 2020

PHXI04:MOTION IN A PLANE

361893 Two particles of equal masses are revolving in circular paths of radii $r_{1}$ and $r_{2}$ respectively with the same speed. The ratio of their centripetal forces is

1

\frac{r_{2}}{r_{1}}

2

\sqrt{\frac{r_{2}}{r_{1}}}

3

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

4

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

Explanation:

$F = \frac{m v^{2}}{r}$ If $m$ and $n$ are constants then
$F \propto \frac{1}{r}$
$∴ \frac{F_{1}}{F_{2}} = (\frac{r_{2}}{r_{1}})$

PHXI04:MOTION IN A PLANE

361894 If a particle covers half the circle of radius $R$ with constant speed $v$ then

1 Change in momentum is

m v r

2 Change in

K

.

E

. is

1 / 2 m v^{2}

3 Change in

K

.

E

. is

m v^{2}

4 Change in

K

.

E

. is zero

Explanation:

As momentum is vector quantity
$∴$ Change in momentum
$Δ P = 2 m v \sin (θ / 2)$
$= 2 m v \sin (90) = 2 m v$
But kinetic energy remains always constant so change in kinetic energy is zero.
supporting img

PHXI04:MOTION IN A PLANE

361891 A wheel is rotating at $900 r p m$ about its axis. When the power is cut off, it comes to rest in $1 \min$ . The angular retardation (in $r a d s^{- 2}$ ) is

1

π / 2

2

π / 4

3

π / 6

4

π / 8

.

Explanation:

As, $ω = ω_{0} + α t$
Here, $ω_{0} = 900 rpm = \frac{(2 π \times 900)}{60} {rads}^{- 1}$ ,
$ω = 0$ and $t = 60 s$
So, $0 = \frac{2 π \times 900}{60} - α \times 60$
$0 = \frac{2 π \times 900}{60} - α \times 60$
$α = \frac{2 π \times 900}{60 \times 60} = \frac{π}{2}$

PHXI04:MOTION IN A PLANE

361892 A particle is moving in a radius $R$ with constant speed $v$ . The magnitude of average acceleration after half revolution is

1

\frac{2 π}{R v}

2

\frac{2 π}{R v^{2}}

3

\frac{2 v^{2}}{π R}

4

\frac{2 v}{π R^{2}}

Explanation:

After half revolution, change in velocity,
$Δ v = v - (- v) = 2 v$
Time taken for half revolution,
$Δ t = \frac{T}{2} = \frac{1}{2} \frac{2 π R}{v} = \frac{π R}{v}$
$∴$ Average acceleration after half revolution,
$a_{a v g} = \frac{Δ v}{Δ t} = \frac{2 v}{(\frac{π R}{v})} = \frac{2 v^{2}}{π R}$

MHTCET - 2020

PHXI04:MOTION IN A PLANE

361893 Two particles of equal masses are revolving in circular paths of radii $r_{1}$ and $r_{2}$ respectively with the same speed. The ratio of their centripetal forces is

1

\frac{r_{2}}{r_{1}}

2

\sqrt{\frac{r_{2}}{r_{1}}}

3

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

4

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

Explanation:

$F = \frac{m v^{2}}{r}$ If $m$ and $n$ are constants then
$F \propto \frac{1}{r}$
$∴ \frac{F_{1}}{F_{2}} = (\frac{r_{2}}{r_{1}})$

PHXI04:MOTION IN A PLANE

361894 If a particle covers half the circle of radius $R$ with constant speed $v$ then

1 Change in momentum is

m v r

2 Change in

K

.

E

. is

1 / 2 m v^{2}

3 Change in

K

.

E

. is

m v^{2}

4 Change in

K

.

E

. is zero

Explanation:

As momentum is vector quantity
$∴$ Change in momentum
$Δ P = 2 m v \sin (θ / 2)$
$= 2 m v \sin (90) = 2 m v$
But kinetic energy remains always constant so change in kinetic energy is zero.
supporting img

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