2022-May-Physics_paper_1__TZ1_SL – All Questions with detailed solution
Q.1.2022-May-Physics_paper_1__TZ1_SL
Topic: Measurements in physics
Discuss:What is the order of magnitude of the wavelength of visible light?
A. $10^{-10} \mathrm{~m}$
B. $ 10^{-7} \mathrm{~m}$
C. $ 10^{-4} \mathrm{~m}$
D. $10^{-1} \mathrm{~m}$
Answer/Explanation
Solution:
Visible light which is detectable by the human eye consists of wavelength ranging from approximately $780 \mathrm{~nm}\left(7.80 \times 10^{-7} \mathrm{~m}\right)-390 \mathrm{~nm}\left(3.90 \times 10^{-7} \mathrm{~m}\right)$. So, the wavelength of light visible to the eye is of the order of $10^{-7} \mathrm{~m}$
Q.2.2022-May-Physics_paper_1__TZ1_SL
Topic: Vector
Given: resultant of two forces acting on a body is 12 N
Calculate: Which pair of forces acting on the body can combine to produce this resultant?
A. $ 1 \mathrm{~N}$ and $2 \mathrm{~N}$
B. $1 \mathrm{~N}$ and $14 \mathrm{~N}$
C. $5 \mathrm{~N}$ and $6 \mathrm{~N}$
D. $6 \mathrm{~N}$ and $7 \mathrm{~N}$
Answer/Explanation
Solution:
If $\mathrm{A}$ and $\mathrm{B}$ are two vector then $R_{\text {resultant }}$ of these will vary from $(A-B)$ to $(A+B)$
$$
\begin{aligned}
& (F)_{\max }=A+B \\
& (F)_{\min }=A-B
\end{aligned}
$$
Option (A)- $1 N$ and $2 N » F_{\max }=3 N, F_{\min }=1 N$
Option (B)- $1 N$ and $14 N » F_{\max }=15 N, F_{\min }=13 N$
Option (C)- $5 N$ and $6 N » F_{\max }=11 N, F_{\min }=1 N$
Option (D)- $6 N$ and $7 N » F_{\max }=13 N, F_{\min }=1 N$
| Hence $12 N$ will get from $6 N$ and $7 N$ |
Q.3.2022-May-Physics_paper_1__TZ1_SL
Topic: Uncertainties and errors
Given: A student measures the time for 20 oscillations of a pendulum. The experiment is repeated four times. The measurements are:
$$
\begin{aligned}
& 10.45 \mathrm{~s} \\
& 10.30 \mathrm{~s} \\
& 10.70 \mathrm{~s} \\
& 10.55 \mathrm{~s}
\end{aligned}
$$
Calculate: What is the best estimate of the uncertainty in the average time for 20 oscillations?
A. $0.01 \mathrm{~s}$
B. $0.05 \mathrm{~s}$
C. $0.2 \mathrm{~s}$
D. $0.5 \mathrm{~s}$
Answer/Explanation
Solution:
Sum of all observations $=10.45+10.30+10.70+10.55=42.00$
$
\langle\text { Average }\rangle=\frac{42}{4}=10.5
$
Sum of modulation error $=\mid$ Average observation-Given observation $\mid$
$
=|10.5-10.45|=0.5
$
$
=|10.3-10.5|=0.2
$
$
=|10.5-10.7|=0.2
$
$
=|10.55-10.5|=0.05
$
$
0.5+0.2+0.2+0.05=0.95
$
$
=\frac{0.95}{4}=0.2375
$
Q.4.2022-May-Physics_paper_1__TZ1_SL
Topic: Motion
Given: $\mathrm{v}$ is brought to rest, after travelling a distanced, by a frictional force$f$
Calculate: second identical block moving with initial speed u is brought to rest in the same distance d by a frictional force $\frac{f}{2}$ . What is u?
A. $ \mathrm{v}$
B. $\frac{\mathrm{v}}{\sqrt{2}}$
C. $\frac{\mathrm{v}}{2}$
D. $\frac{\mathrm{v}}{4}$
Answer/Explanation
Solution:
Let the mass of block be m ,then acceleration in first case when friction is f,
acceleration, $a=\frac{f}{m}$ from $[\mathrm{F}=\mathrm{ma}]$
$
\begin{aligned}
& v^2=u^2+2 a s \\
& v=\text { finel velocity } \\
& u=\text { initial velocity } \\
& a=\text { acceleration of body } \\
& s=\text { distance travvelled }
\end{aligned}\quad\text{(equation of motion )}
$
$V_f=0$
$V_i=V$
Using eq. of motion,
$0^2=V^2-2(\frac{f}{m})d$
$
\left(a d=\frac{V^2}{2}\right)
$
In case 2 , when friction is $\mathrm{f} / 2$ aceeleration will be $\frac{f}{2 m}$, which is half of previous acceleration,
$
\begin{aligned}
& u^2=\frac{2 a}{2 d}=a d \quad\left(a d=\frac{V^2}{2}\right) \\
& \therefore u=\frac{V}{\sqrt{2}}
\end{aligned}
$
| $u=\frac{V}{\sqrt{2}}$ |