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PHXI05:LAWS OF MOTION

363467 If a street light of mass $M$ is suspended from the end of a uniform rod of length $L$ in different possible patterns as shown in figure, then
supporting img

1 Pattern

A

is more sturdy

2 Pattern

B

is more sturdy

3 Pattern

C

is more sturdy

4 All will have same sturdiness

Explanation:

For equilibrium of street light, net torque on rod should be zero.
$m g \times x = T \times y$
or $T = \frac{m g x}{y}$
supporting img
For $T$ to be minimum, $y$ should be maximum.
Hence, pattern $A$ is more study.

AIIMS - 2006

PHXI05:LAWS OF MOTION

363468 A body of mass $500 g$ moves along $x$ -axis such that it's velocity varies with displacement $x$ according to the relation $v = 10 \sqrt{x} m / s$ , the force acting on the body is

1

5 N

2

125 N

3

166 N

4

25 N

Explanation:

Mass, $m = 500 g = 0.5 k g$
velocity $v = 10 \sqrt{x} m / s$
$a = \frac{d \vec{v}}{d t} = \frac{d \vec{v}}{d x} \cdot (\frac{d x}{d t}) = \frac{1}{2} (10) x^{- 1 / 2} \frac{d x}{d t}$
$= \frac{1}{2} \frac{(10)}{\sqrt{x}} \cdot v = \frac{1}{2} \frac{(10)}{\sqrt{x}} 10 (\sqrt{x});$
$a = 50 m / s^{2}$
Force, $F = m a = 0.5 \times 50 = 25 N$
$∴ \vec{F} = 25 \hat{i} N .$

JEE - 2023

PHXI05:LAWS OF MOTION

363469 Assertion :
A man in a closed cabin falling freely does not experience gravity.
Reason :
Inertia and gravitational mass have equivalence.

1 Both assertion and reason are correct and reason is the correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$m_{g r a v .} g - N = m_{i n e r t i a l} . a$ For freely falling
$a = g$ . Since $m_{g r a v} = m_{i n e r t} \Rightarrow N = 0$

PHXI05:LAWS OF MOTION

363470 Two forces having magnitude $A$ and $\frac{A}{2}$ are perpendicular to each other. The magnitude of their resultant is

1

\frac{5 A}{2}

2

\frac{\sqrt{5} A^{2}}{2}

3

\frac{\sqrt{5} A}{4}

4

\frac{\sqrt{5} A}{2}

Explanation:

Force is vector quantity where resultant of two forces is given as
By parallelogram Law,
$F^{2} = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos θ$
$= (A)^{2} + {(\frac{A}{2})}^{2} + 2 \times A \times \frac{A}{2} \times \cos 90^{\circ}$
(As forces are acting perpendicularly)
$F^{2} = \frac{5 A^{2}}{4} \Rightarrow F = \frac{\sqrt{5} A}{2}$ units

JEE - 2023

PHXI05:LAWS OF MOTION

363471 Assertion :
From Newton's second law of motion impulse is equal to change in momentum.
Reason :
Impulse and momentum have different dimensions.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Impulse $= F (Δ t) = Δ p$
Since impulse equals the change in momentum, impulse and momentum have the same dimensions. So correct option is (3)

PHXI05:LAWS OF MOTION

363467 If a street light of mass $M$ is suspended from the end of a uniform rod of length $L$ in different possible patterns as shown in figure, then
supporting img

1 Pattern

A

is more sturdy

2 Pattern

B

is more sturdy

3 Pattern

C

is more sturdy

4 All will have same sturdiness

Explanation:

For equilibrium of street light, net torque on rod should be zero.
$m g \times x = T \times y$
or $T = \frac{m g x}{y}$
supporting img
For $T$ to be minimum, $y$ should be maximum.
Hence, pattern $A$ is more study.

AIIMS - 2006

PHXI05:LAWS OF MOTION

363468 A body of mass $500 g$ moves along $x$ -axis such that it's velocity varies with displacement $x$ according to the relation $v = 10 \sqrt{x} m / s$ , the force acting on the body is

1

5 N

2

125 N

3

166 N

4

25 N

Explanation:

Mass, $m = 500 g = 0.5 k g$
velocity $v = 10 \sqrt{x} m / s$
$a = \frac{d \vec{v}}{d t} = \frac{d \vec{v}}{d x} \cdot (\frac{d x}{d t}) = \frac{1}{2} (10) x^{- 1 / 2} \frac{d x}{d t}$
$= \frac{1}{2} \frac{(10)}{\sqrt{x}} \cdot v = \frac{1}{2} \frac{(10)}{\sqrt{x}} 10 (\sqrt{x});$
$a = 50 m / s^{2}$
Force, $F = m a = 0.5 \times 50 = 25 N$
$∴ \vec{F} = 25 \hat{i} N .$

JEE - 2023

PHXI05:LAWS OF MOTION

363469 Assertion :
A man in a closed cabin falling freely does not experience gravity.
Reason :
Inertia and gravitational mass have equivalence.

1 Both assertion and reason are correct and reason is the correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$m_{g r a v .} g - N = m_{i n e r t i a l} . a$ For freely falling
$a = g$ . Since $m_{g r a v} = m_{i n e r t} \Rightarrow N = 0$

PHXI05:LAWS OF MOTION

363470 Two forces having magnitude $A$ and $\frac{A}{2}$ are perpendicular to each other. The magnitude of their resultant is

1

\frac{5 A}{2}

2

\frac{\sqrt{5} A^{2}}{2}

3

\frac{\sqrt{5} A}{4}

4

\frac{\sqrt{5} A}{2}

Explanation:

Force is vector quantity where resultant of two forces is given as
By parallelogram Law,
$F^{2} = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos θ$
$= (A)^{2} + {(\frac{A}{2})}^{2} + 2 \times A \times \frac{A}{2} \times \cos 90^{\circ}$
(As forces are acting perpendicularly)
$F^{2} = \frac{5 A^{2}}{4} \Rightarrow F = \frac{\sqrt{5} A}{2}$ units

JEE - 2023

PHXI05:LAWS OF MOTION

363471 Assertion :
From Newton's second law of motion impulse is equal to change in momentum.
Reason :
Impulse and momentum have different dimensions.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Impulse $= F (Δ t) = Δ p$
Since impulse equals the change in momentum, impulse and momentum have the same dimensions. So correct option is (3)

PHXI05:LAWS OF MOTION

363467 If a street light of mass $M$ is suspended from the end of a uniform rod of length $L$ in different possible patterns as shown in figure, then
supporting img

1 Pattern

A

is more sturdy

2 Pattern

B

is more sturdy

3 Pattern

C

is more sturdy

4 All will have same sturdiness

Explanation:

For equilibrium of street light, net torque on rod should be zero.
$m g \times x = T \times y$
or $T = \frac{m g x}{y}$
supporting img
For $T$ to be minimum, $y$ should be maximum.
Hence, pattern $A$ is more study.

AIIMS - 2006

PHXI05:LAWS OF MOTION

363468 A body of mass $500 g$ moves along $x$ -axis such that it's velocity varies with displacement $x$ according to the relation $v = 10 \sqrt{x} m / s$ , the force acting on the body is

1

5 N

2

125 N

3

166 N

4

25 N

Explanation:

Mass, $m = 500 g = 0.5 k g$
velocity $v = 10 \sqrt{x} m / s$
$a = \frac{d \vec{v}}{d t} = \frac{d \vec{v}}{d x} \cdot (\frac{d x}{d t}) = \frac{1}{2} (10) x^{- 1 / 2} \frac{d x}{d t}$
$= \frac{1}{2} \frac{(10)}{\sqrt{x}} \cdot v = \frac{1}{2} \frac{(10)}{\sqrt{x}} 10 (\sqrt{x});$
$a = 50 m / s^{2}$
Force, $F = m a = 0.5 \times 50 = 25 N$
$∴ \vec{F} = 25 \hat{i} N .$

JEE - 2023

PHXI05:LAWS OF MOTION

363469 Assertion :
A man in a closed cabin falling freely does not experience gravity.
Reason :
Inertia and gravitational mass have equivalence.

1 Both assertion and reason are correct and reason is the correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$m_{g r a v .} g - N = m_{i n e r t i a l} . a$ For freely falling
$a = g$ . Since $m_{g r a v} = m_{i n e r t} \Rightarrow N = 0$

PHXI05:LAWS OF MOTION

363470 Two forces having magnitude $A$ and $\frac{A}{2}$ are perpendicular to each other. The magnitude of their resultant is

1

\frac{5 A}{2}

2

\frac{\sqrt{5} A^{2}}{2}

3

\frac{\sqrt{5} A}{4}

4

\frac{\sqrt{5} A}{2}

Explanation:

Force is vector quantity where resultant of two forces is given as
By parallelogram Law,
$F^{2} = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos θ$
$= (A)^{2} + {(\frac{A}{2})}^{2} + 2 \times A \times \frac{A}{2} \times \cos 90^{\circ}$
(As forces are acting perpendicularly)
$F^{2} = \frac{5 A^{2}}{4} \Rightarrow F = \frac{\sqrt{5} A}{2}$ units

JEE - 2023

PHXI05:LAWS OF MOTION

363471 Assertion :
From Newton's second law of motion impulse is equal to change in momentum.
Reason :
Impulse and momentum have different dimensions.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Impulse $= F (Δ t) = Δ p$
Since impulse equals the change in momentum, impulse and momentum have the same dimensions. So correct option is (3)

PHXI05:LAWS OF MOTION

363467 If a street light of mass $M$ is suspended from the end of a uniform rod of length $L$ in different possible patterns as shown in figure, then
supporting img

1 Pattern

A

is more sturdy

2 Pattern

B

is more sturdy

3 Pattern

C

is more sturdy

4 All will have same sturdiness

Explanation:

For equilibrium of street light, net torque on rod should be zero.
$m g \times x = T \times y$
or $T = \frac{m g x}{y}$
supporting img
For $T$ to be minimum, $y$ should be maximum.
Hence, pattern $A$ is more study.

AIIMS - 2006

PHXI05:LAWS OF MOTION

363468 A body of mass $500 g$ moves along $x$ -axis such that it's velocity varies with displacement $x$ according to the relation $v = 10 \sqrt{x} m / s$ , the force acting on the body is

1

5 N

2

125 N

3

166 N

4

25 N

Explanation:

Mass, $m = 500 g = 0.5 k g$
velocity $v = 10 \sqrt{x} m / s$
$a = \frac{d \vec{v}}{d t} = \frac{d \vec{v}}{d x} \cdot (\frac{d x}{d t}) = \frac{1}{2} (10) x^{- 1 / 2} \frac{d x}{d t}$
$= \frac{1}{2} \frac{(10)}{\sqrt{x}} \cdot v = \frac{1}{2} \frac{(10)}{\sqrt{x}} 10 (\sqrt{x});$
$a = 50 m / s^{2}$
Force, $F = m a = 0.5 \times 50 = 25 N$
$∴ \vec{F} = 25 \hat{i} N .$

JEE - 2023

PHXI05:LAWS OF MOTION

363469 Assertion :
A man in a closed cabin falling freely does not experience gravity.
Reason :
Inertia and gravitational mass have equivalence.

1 Both assertion and reason are correct and reason is the correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$m_{g r a v .} g - N = m_{i n e r t i a l} . a$ For freely falling
$a = g$ . Since $m_{g r a v} = m_{i n e r t} \Rightarrow N = 0$

PHXI05:LAWS OF MOTION

363470 Two forces having magnitude $A$ and $\frac{A}{2}$ are perpendicular to each other. The magnitude of their resultant is

1

\frac{5 A}{2}

2

\frac{\sqrt{5} A^{2}}{2}

3

\frac{\sqrt{5} A}{4}

4

\frac{\sqrt{5} A}{2}

Explanation:

Force is vector quantity where resultant of two forces is given as
By parallelogram Law,
$F^{2} = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos θ$
$= (A)^{2} + {(\frac{A}{2})}^{2} + 2 \times A \times \frac{A}{2} \times \cos 90^{\circ}$
(As forces are acting perpendicularly)
$F^{2} = \frac{5 A^{2}}{4} \Rightarrow F = \frac{\sqrt{5} A}{2}$ units

JEE - 2023

PHXI05:LAWS OF MOTION

363471 Assertion :
From Newton's second law of motion impulse is equal to change in momentum.
Reason :
Impulse and momentum have different dimensions.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Impulse $= F (Δ t) = Δ p$
Since impulse equals the change in momentum, impulse and momentum have the same dimensions. So correct option is (3)

PHXI05:LAWS OF MOTION

363467 If a street light of mass $M$ is suspended from the end of a uniform rod of length $L$ in different possible patterns as shown in figure, then
supporting img

1 Pattern

A

is more sturdy

2 Pattern

B

is more sturdy

3 Pattern

C

is more sturdy

4 All will have same sturdiness

Explanation:

For equilibrium of street light, net torque on rod should be zero.
$m g \times x = T \times y$
or $T = \frac{m g x}{y}$
supporting img
For $T$ to be minimum, $y$ should be maximum.
Hence, pattern $A$ is more study.

AIIMS - 2006

PHXI05:LAWS OF MOTION

363468 A body of mass $500 g$ moves along $x$ -axis such that it's velocity varies with displacement $x$ according to the relation $v = 10 \sqrt{x} m / s$ , the force acting on the body is

1

5 N

2

125 N

3

166 N

4

25 N

Explanation:

Mass, $m = 500 g = 0.5 k g$
velocity $v = 10 \sqrt{x} m / s$
$a = \frac{d \vec{v}}{d t} = \frac{d \vec{v}}{d x} \cdot (\frac{d x}{d t}) = \frac{1}{2} (10) x^{- 1 / 2} \frac{d x}{d t}$
$= \frac{1}{2} \frac{(10)}{\sqrt{x}} \cdot v = \frac{1}{2} \frac{(10)}{\sqrt{x}} 10 (\sqrt{x});$
$a = 50 m / s^{2}$
Force, $F = m a = 0.5 \times 50 = 25 N$
$∴ \vec{F} = 25 \hat{i} N .$

JEE - 2023

PHXI05:LAWS OF MOTION

363469 Assertion :
A man in a closed cabin falling freely does not experience gravity.
Reason :
Inertia and gravitational mass have equivalence.

1 Both assertion and reason are correct and reason is the correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$m_{g r a v .} g - N = m_{i n e r t i a l} . a$ For freely falling
$a = g$ . Since $m_{g r a v} = m_{i n e r t} \Rightarrow N = 0$

PHXI05:LAWS OF MOTION

363470 Two forces having magnitude $A$ and $\frac{A}{2}$ are perpendicular to each other. The magnitude of their resultant is

1

\frac{5 A}{2}

2

\frac{\sqrt{5} A^{2}}{2}

3

\frac{\sqrt{5} A}{4}

4

\frac{\sqrt{5} A}{2}

Explanation:

Force is vector quantity where resultant of two forces is given as
By parallelogram Law,
$F^{2} = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos θ$
$= (A)^{2} + {(\frac{A}{2})}^{2} + 2 \times A \times \frac{A}{2} \times \cos 90^{\circ}$
(As forces are acting perpendicularly)
$F^{2} = \frac{5 A^{2}}{4} \Rightarrow F = \frac{\sqrt{5} A}{2}$ units

JEE - 2023

PHXI05:LAWS OF MOTION

363471 Assertion :
From Newton's second law of motion impulse is equal to change in momentum.
Reason :
Impulse and momentum have different dimensions.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Impulse $= F (Δ t) = Δ p$
Since impulse equals the change in momentum, impulse and momentum have the same dimensions. So correct option is (3)