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2022-Nov-Physics_paper_1__TZ0_SL – All Questions with detailed solution

Q.1.2022-Nov-Physics_paper_1__TZ0_SL

Topic: Uncertainties and errors

Given: Rectangular sheet of paper has dimensions of $(30.0 \pm 0.5) \mathrm{cm}$ and $(20.0 \pm 0.5) \mathrm{cm}$.

Calculate: What is the percentage uncertainty of the perimeter of the paper?

A. $1 \%$

B. $2 \%$

C. $2.5 \%$

D. $4 \%$

Answer/Explanation

Solution:

The perimeter $P$ of the rectangular sheet of paper is given by:

$P = 2(l+w)$

where $l$ and $w$ are the length and width of the rectangle, respectively.

Using the values given, we have:

$l = (30.0 \pm 0.5) \mathrm{cm}$$ $$w = (20.0 \pm 0.5) \mathrm{cm}$

So the perimeter is:

$P = 2[(30.0) \mathrm{cm} + (20.0 ) \mathrm{cm}]$

Simplifying:

$P = 2 \times 50.0 \mathrm{cm} = 100.0 \mathrm{cm}$

Therefore, the absolute uncertainty in the perimeter is:

$\Delta P = 2(\Delta \mathrm{0.5}+\Delta \mathrm{0.5})=2 \mathrm{cm}$

And the percentage uncertainty is:

$\frac{\Delta P}{P} \times 100\%=\frac{2.0 \mathrm{cm}}{100.0 \mathrm{cm}} \times 100\% = 2.0\%$

Therefore, the percentage uncertainty of the perimeter of the paper is $\boxed{2.0\%}$.

$\colorbox{yellow}{Correct Option -B}$

Q.2.2022-Nov-Physics_paper_1__TZ0_SL

Topic: Vectors and scalars

Given: Two forces, $\mathrm{F}$ and $\mathrm{G}$, act on a system.

Now, $\mathrm{F}$ is reversed in direction and $\mathrm{G}$ is halved.

Discuss: Which vector correctly represents the new resultant force?

Answer/Explanation

Solution:

When the direction of $\mathrm{F}$ is reversed and the magnitude of $\mathrm{G}$ is halved.

$\colorbox{yellow}{Correct Option -D}$

Q.3.2022-Nov-Physics_paper_1__TZ0_SL

Topic: Motion

Given: Ball 1 is dropped from rest from an initial height $h.$ At the same instant, ball 2 is launched vertically upwards at an initial velocity $u$.

Calculate: At what time are both balls at the same distance above the ground?

A. $\frac{h}{4 u}$

B. $\frac{h}{2 u}$

C. $\frac{h}{u}$

D. $\frac{2 h}{u}$

Answer/Explanation

Solution:

Let both balls meet at $\mathrm{H}$

Using equation of motion for ball 1 which is launched vertically up with velocity $u$,

\( H=u t-\frac{1}{2} gt^2\quad\text{[eq-1]} \)

Now, writing the eqaution of motion for ball 2 which is dropped ($u=0$) in this case,

\(h- H=\frac{1}{2} gt^2\quad\text{[eq-2]} \)

$\text{Adding eq-(1) and eq-(2)}$

$h=ut$

$t=\frac{h}{u}$

$\colorbox{yellow}{Correct Option -C}$

Q.4.2022-Nov-Physics_paper_1__TZ0_SL

Topic: Motion

Given: A projectile is launched with a velocity $u$ at an angle $\theta$ to the horizontal. It reaches a maximum height $s$.

Calculate: What is the time taken to reach the maximum height?

A. $\frac{2 s}{u \cos \theta}$
B. $\frac{2 s}{g}$
C. $\frac{u \cos \theta}{g}$
D. $\frac{u \sin \theta}{g}$

Answer/Explanation

Solution:

The time taken for a projectile to reach its maximum height :

$t = \frac{u \sin\theta}{g}$

where $u$ is the initial velocity, $\theta$ is the angle of projection, $g$ is the acceleration due to gravity, and $t$ is the time taken to reach the maximum height.

To derive this formula, we use the fact that the vertical component of velocity at the maximum height is zero, i.e., $v_y = u \sin\theta – gt = 0$, where $v_y$ is the vertical component of velocity. Solving for $t$, we get $t = \frac{u \sin\theta}{g}$.

Therefore, the time taken to reach the maximum height is $\boxed{t = \frac{u \sin\theta}{g}}$.

$\colorbox{yellow}{Correct Option -D}$
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