12. a uniform half-metre rule ab is balanced horizontally on a knife edge placed 20 cm from a and with a mass of 30 g at a. calculate the mass of the rule.
a. 20 g
b. 30 g
c. 120 g
d. 160 g

13. a body executing a simple harmonic motion has an angular speed of 2 rads⁻¹ and maximum displacement of 5 cm. determine the magnitude of its acceleration.
a. 0.1 m s⁻²
b. 0.2 m s⁻²
c. 0.4 m s⁻²
d. 0.5 m s⁻²

14. a force acts on a body of mass 20 kg and changes its speed from 20 to 30 m s⁻¹. calculate the magnitude of the impulse.
a. 200 n s
b. 400 n s
c. 600 n s
d. 1000 n s

15. the absolute zero is the temperature at which
a. water freezes.
b. molecular activity ceases.
c. the states of matter co-exist in equilibrium.
d. water has minimum volume.

16. which of the following sources of energy is non-renewable?
a. petroleum
b. sun
c. wind
d. tidal waves

17. the moment of a force has the same dimensions as those of
a. efficiency.
b. momentum.
c. power.
d. work.

18. a body 150 kg is raised through a vertical height of 5 m in 100 s by a machine. if the efficiency of the machine is 80 %, calculate the input power. g = 10 m s⁻².
a. 9.60 w
b. 60.00 w
c. 75.00 w
d. 93.75 w

Question

12. a uniform half-metre rule ab is balanced horizontally on a knife edge placed 20 cm from a and with a mass of 30 g at a. calculate the mass of the rule.
a. 20 g
b. 30 g
c. 120 g
d. 160 g

13. a body executing a simple harmonic motion has an angular speed of 2 rads⁻¹ and maximum displacement of 5 cm. determine the magnitude of its acceleration.
a. 0.1 m s⁻²
b. 0.2 m s⁻²
c. 0.4 m s⁻²
d. 0.5 m s⁻²

14. a force acts on a body of mass 20 kg and changes its speed from 20 to 30 m s⁻¹. calculate the magnitude of the impulse.
a. 200 n s
b. 400 n s
c. 600 n s
d. 1000 n s

15. the absolute zero is the temperature at which
a. water freezes.
b. molecular activity ceases.
c. the states of matter co-exist in equilibrium.
d. water has minimum volume.

16. which of the following sources of energy is non-renewable?
a. petroleum
b. sun
c. wind
d. tidal waves

17. the moment of a force has the same dimensions as those of
a. efficiency.
b. momentum.
c. power.
d. work.

18. a body 150 kg is raised through a vertical height of 5 m in 100 s by a machine. if the efficiency of the machine is 80 %, calculate the input power. g = 10 m s⁻².
a. 9.60 w
b. 60.00 w
c. 75.00 w
d. 93.75 w

Accepted Answer

### Question 12
# Explanation:
## Step1: Identify pivot and moments
The half - metre rule has length $L = 50\space cm$. The pivot (knife - edge) is at $20\space cm$ from $A$. The mass at $A$ is $m_1=30\space g$ and let the mass of the rule be $M$. The center of mass of the uniform rule is at its mid - point, i.e., at $25\space cm$ from $A$. The distance of the mass at $A$ from the pivot, $d_1 = 20\space cm$ and the distance of the center of mass of the rule from the pivot, $d_2=25 - 20=5\space cm$.

Taking moments about the pivot, the clockwise moment equals the anticlockwise moment. The moment due to the mass at $A$ is $m_1\times g\times d_1$ (anticlockwise) and the moment due to the mass of the rule is $M\times g\times d_2$ (clockwise). Since $g$ is common on both sides, we can write $m_1\times d_1=M\times d_2$.

## Step2: Solve for $M$
We know that $m_1 = 30\space g$, $d_1 = 20\space cm$ and $d_2 = 5\space cm$. Substituting these values into the equation $m_1\times d_1=M\times d_2$, we get:

$30\times20 = M\times5$

$M=\frac{30\times20}{5}$

$M = 120\space g$

# Answer:
C. 120 g

### Question 13
# Explanation:
## Step1: Recall the formula for acceleration in SHM
The formula for the acceleration $a$ of a body in simple harmonic motion is $a=\omega^{2}x$, where $\omega$ is the angular speed and $x$ is the displacement.

## Step2: Substitute the values
Given that $\omega = 2\space rad/s$ and $x = 5\space cm=0.05\space m$.

Substitute these values into the formula:

$a=(2)^{2}\times0.05$

$a = 4\times0.05$

$a=0.2\space m/s^{2}$

# Answer:
B. $0.2\space m/s^{2}$

### Question 14
# Explanation:
## Step1: Recall the formula for impulse
Impulse $J$ is equal to the change in momentum $\Delta p$. The formula for momentum is $p = mv$, so $\Delta p=m(v_2 - v_1)$, where $m$ is the mass, $v_2$ is the final velocity and $v_1$ is the initial velocity.

## Step2: Substitute the values
Given that $m = 20\space kg$, $v_1 = 20\space m/s$ and $v_2 = 30\space m/s$.

$\Delta p=20\times(30 - 20)$

$\Delta p=20\times10$

$\Delta p = 200\space Ns$

# Answer:
A. 200 N s

### Question 15
# Explanation:
Absolute zero is the lowest possible temperature. At absolute zero ( $0\space K$ or $- 273.15^{\circ}C$ ), the molecular motion (kinetic energy of molecules) ceases. Water freezes at $0^{\circ}C$ (not absolute zero). The states of matter co - exist at the triple point of water. The volume of water is minimum at $4^{\circ}C$ (anomalous expansion of water).

# Brief Explanations:
Absolute zero is the temperature at which molecular activity (motion) ceases. Option A is wrong (water freezes at $0^{\circ}C$), option C is wrong (states co - exist at triple point), option D is wrong (water has minimum volume at $4^{\circ}C$).
# Answer:
B. molecular activity ceases.

### Question 16
# Explanation:
Non - renewable energy sources are those that are finite and take a very long time to be replenished. Petroleum is a fossil fuel formed over millions of years from the remains of dead organisms. Sun, wind, and tidal waves are renewable energy sources as they are continuously available.

# Brief Explanations:
Petroleum is a non - renewable energy source. Sun, wind, and tidal waves are renewable.
# Answer:
A. Petroleum

### Question 17
# Explanation:
The moment of a force $M = F\times d$, where $F$ has dimensions $[MLT^{- 2}]$ and $d$ has dimensions $[L]$, so the dimension of moment of force is $[ML^{2}T^{-2}]$.

- Efficiency is dimensionless.
- Momentum $p = mv$, dimension of $m$ is $[M]$ and $v$ is $[LT^{-1}]$, so momentum has dimension $[MLT^{-1}]$.
- Power $P=\frac{W}{t}$, work $W = F\times d$ has dimension $[ML^{2}T^{-2}]$ and $t$ has dimension $[T]$, so power has dimension $[ML^{2}T^{-3}]$.
- Work $W=F\times d$, dimension of $F$ is $[MLT^{-2}]$ and $d$ is $[L]$, so work has dimension $[ML^{2}T^{-2}]$, same as the moment of a force.

# Brief Explanations:
Moment of force ($F\times d$) has dimension $[ML^{2}T^{-2}]$, same as work ($F\times d$). Efficiency is dimensionless, momentum has $[MLT^{-1}]$, power has $[ML^{2}T^{-3}]$.
# Answer:
D. work.

### Question 18
# Explanation:
## Step1: Calculate the useful work done
The useful work done $W_{useful}$ in raising the body is given by $W_{useful}=mgh$, where $m = 150\space kg$, $g = 10\space m/s^{2}$ and $h = 5\space m$.

$W_{useful}=150\times10\times5=7500\space J$

## Step2: Calculate the useful power
Useful power $P_{useful}=\frac{W_{useful}}{t}$, where $t = 100\space s$.

$P_{useful}=\frac{7500}{100}=75\space W$

## Step3: Calculate the input power
Efficiency $\eta=\frac{P_{useful}}{P_{input}}\times100$. Given $\eta = 80\%=0.8$

$0.8=\frac{75}{P_{input}}$

$P_{input}=\frac{75}{0.8}=93.75\space W$

# Answer:
D. 93.75 W

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QUESTION IMAGE

Question

Explanation:

Question 12

Step1: Identify pivot and moments

Step2: Solve for \(M\)

Step1: Recall the formula for acceleration in SHM

Step2: Substitute the values

Step1: Recall the formula for impulse

Step2: Substitute the values

Answer:

Question 13