3. hot liquid in a vacuum flask cools extremely slowly because some modes of heat transfer cannot take place in a vacuum. these modes are
a. conduction and convection only.
b. conduction and radiation only.
c. conduction, convection and radiation.
d. convection and radiation only.

4. a swimmer coming out of a pool of water soon feels cold on a sunny day. the swimmer experienced coldness because
a. water evaporates from the skin.
b. convection occurs in the air.
c. air is a bad conductor of heat.
d. water is a good conductor of heat.

5. a car of mass 1000 kg travelling at 72 km hr⁻¹ is brought to rest over a distance of 30 m. calculate the retardation of the car.
a. 7.1 ms⁻²
b. 6.2 ms⁻²
c. 6.7 ms⁻²
d. 5.8 ms⁻²

6. which of the following actions is a process of converting a moving-coil galvanometer into an ammeter? connecting a resistor of
a. large resistance in series with it.
b. small resistance in parallel with it.
c. small resistance in series with it.
d. large resistance in parallel with it.

7. a ball, x is dropped from a height of 200 m at the same instance that another ball, y is projected vertically from the ground with an initial velocity of 100 ms⁻¹. at what height above the ground will the two balls meet? g = 10 ms⁻²
a. 180 m
b. 100 m
c. 20 m
d. 220 m

Question

3. hot liquid in a vacuum flask cools extremely slowly because some modes of heat transfer cannot take place in a vacuum. these modes are
a. conduction and convection only.
b. conduction and radiation only.
c. conduction, convection and radiation.
d. convection and radiation only.

4. a swimmer coming out of a pool of water soon feels cold on a sunny day. the swimmer experienced coldness because
a. water evaporates from the skin.
b. convection occurs in the air.
c. air is a bad conductor of heat.
d. water is a good conductor of heat.

5. a car of mass 1000 kg travelling at 72 km hr⁻¹ is brought to rest over a distance of 30 m. calculate the retardation of the car.
a. 7.1 ms⁻²
b. 6.2 ms⁻²
c. 6.7 ms⁻²
d. 5.8 ms⁻²

6. which of the following actions is a process of converting a moving-coil galvanometer into an ammeter? connecting a resistor of
a. large resistance in series with it.
b. small resistance in parallel with it.
c. small resistance in series with it.
d. large resistance in parallel with it.

7. a ball, x is dropped from a height of 200 m at the same instance that another ball, y is projected vertically from the ground with an initial velocity of 100 ms⁻¹. at what height above the ground will the two balls meet? g = 10 ms⁻²
a. 180 m
b. 100 m
c. 20 m
d. 220 m

Accepted Answer

### Question 3
# Brief Explanations:
Heat transfer modes: Conduction needs a medium (molecules collide), convection needs a fluid medium (movement of fluid), radiation is electromagnetic (no medium). Vacuum has no medium, so conduction and convection can't occur, but radiation can. So the modes that can't take place in vacuum are conduction and convection.
# Answer:
A. conduction and convection only.

### Question 4
# Brief Explanations:
When a swimmer comes out, water on skin evaporates. Evaporation is a cooling process (requires heat, taken from skin). Convection in air or air's conductivity isn't the main reason here, and water is a poor conductor. So the coldness is due to water evaporation from skin.
# Answer:
A. water evaporates from the skin.

### Question 5
# Explanation:
## Step1: Convert speed to m/s
Initial speed $ u = 72\space km/h=\frac{72\times1000}{3600}=20\space m/s $, final speed $ v = 0\space m/s $, distance $ s = 30\space m $.
## Step2: Use $ v^{2}=u^{2}+2as $
Rearrange for acceleration $ a=\frac{v^{2}-u^{2}}{2s} $. Substitute values: $ a=\frac{0 - 20^{2}}{2\times30}=\frac{- 400}{60}\approx - 6.7\space m/s^{2} $ (retardation is magnitude, so $ 6.7\space m/s^{2} $).
# Answer:
C. $ 6.7\space ms^{-2} $

### Question 6
# Brief Explanations:
To convert a galvanometer to ammeter, a shunt (low resistance) is connected in parallel. This allows most current to pass through the shunt, so the galvanometer can measure larger currents. Series with large resistance is for voltmeter, parallel with large resistance or series with small resistance don't work. So connect small resistance in parallel.
# Answer:
B. small resistance in parallel with it.

### Question 7
# Explanation:
## Step1: Equations of motion for both balls
For ball X (dropped, $ u_X = 0 $): $ h_X=200-\frac{1}{2}gt^{2} $ (distance fallen from 200m, so height above ground is $ 200 - h_{fallen} $).
For ball Y (projected up, $ u_Y = 100\space m/s $): $ h_Y=u_Yt-\frac{1}{2}gt^{2} $.
When they meet, $ h_X = h_Y $, so $ 200-\frac{1}{2}gt^{2}=100t-\frac{1}{2}gt^{2} $.
## Step2: Solve for time $ t $
The $ \frac{1}{2}gt^{2} $ terms cancel: $ 200 = 100t \Rightarrow t = 2\space s $.
## Step3: Find height $ h_Y $ (or $ h_X $)
Substitute $ t = 2 $ into $ h_Y $: $ h_Y=100\times2-\frac{1}{2}\times10\times2^{2}=200 - 20 = 180\space m $.
# Answer:
A. $ 180\space m $

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QUESTION IMAGE

Question

Explanation:

Question 3

Step1: Convert speed to m/s

Step2: Use \( v^{2}=u^{2}+2as \)

Answer:

Question 4