SECTION - B [PHYSICS - NUMERIC] - JEE STD 11 Science Questions

Question 14 Marks

A particle is projected at an angle of $30^{\circ}$ from horizontal at a speed of $60 m / s$. The height traversed by the particle in the first second is $h _0$ and height traversed in the last second, before it reaches the maximum height, is $h_1$. The ratio $h_0: h_1$ is ___________
[Take, $g =10 m / s ^2$ ]

Answer

(5)
Sol.

$\begin{array}{l} S _1=30 \times 1-\frac{1}{2} \times 10 \times 1=25 \\ S_3=30+\left(\frac{-10}{2}\right) \times(2 \times 3-1)=5 \\ \frac{S_1}{S_3}=\frac{25}{5}=5\end{array}$

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Question 24 Marks

The position vectors of two 1 kg particles, (A) and (B), are given by$
\overrightarrow{r}_{A}=\left(\alpha_1 t^2 \hat{i}+\alpha_2 \hat{t}+\alpha_3 t \hat{k}\right) m
$ and $\overrightarrow{ r }_{ B }=\left(\beta_1 t \hat{ i }+\beta_2 t ^2 \hat{ j }+\beta_3 t \hat{ k }\right) m$, respectively;
$\left(\alpha_1=1 m / s ^2, \alpha_2=3 n m / s , \alpha_3=2 m / s , \beta_1=2 m / s\right.$, $\beta_2=-1 m / s ^2, \beta_3=4 p m / s$ ), where t is time, n and p are constants, At $t=1 s,\left|\overrightarrow{ V }_{ A }\right|=\left|\overrightarrow{ V }_{ B }\right|$ and velocities $\vec{V}_A$ and $\vec{V}_B$ of the particles are orthogonal to each other. At $t=1 s$, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is $\sqrt{ L } kgm ^2 s^{-1}$. The value of $L$ is _________ .

Answer

(90)
$
\begin{array}{l}Sol.\
V_A=(2 t \hat{i}+3 n \hat{j}+2 \hat{k}) \\
\vec{V}_B=(2 \hat{i}-2 \hat{t} \hat{j}+4 p \hat{k}) \\
\overrightarrow{V}_{A} \cdot \overrightarrow{V}_{B}=0 \\
4-6 n+8 p=0 \\
2-3 n+4 p=0 \\
3 n=2+4 p \\
\left|\overrightarrow{V}_{A}\right|=\left|\overrightarrow{V}_{B}\right| \\
4+9 n^2+4=4+4+16 p^2 \\
p=\frac{-1}{4} \quad \Rightarrow n=\frac{1}{3} \\
\overrightarrow{L}=m_{A}\left(\overrightarrow{r}_{A \cdot B} \times \overrightarrow{V}_{A}\right) \\
\overrightarrow{r}_{A / B}=\left(\alpha_1-\beta_1\right) \hat{i}+\left(\alpha_2-\beta_2\right) \hat{j}+\left(\alpha_3-\beta_3\right) \\
=(1-2) \hat{i}+(1+1) \hat{j}+3 \hat{k} \\
=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 2 & 3 \\
2 & 1 & 2
\end{array}\right|=\hat{i}+8 \hat{j}-5 \hat{k} \\
=\sqrt{1+64+25}=\sqrt{90}
\end{array}
$

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Question 34 Marks

Three conductions of same length having thermal conductivity $k _1, k _2$ and $k _3$ are connected as shown in figure.

Area of cross sections of $1^{14}$ and $2^{\text {rad }}$ conductor are same and for $3^{\text {nd }}$ conductor it is double of the $1^st$ conductor. The temperatures are given in the figure. In steady state condition, the value of $\theta$ is _________ ${ }^{\circ} C$.
(Given : $k _1=60 Js ^{-1} m^{-1} K^{-1}, k _2=120 Js ^{-1} m^{-1} K^{-1}, k _3=$ $135 Js ^{-1} m^{-1} K^{-1}$ )

Answer

(40)

$\begin{array}{l}Sol.\ R _1=\frac{2 L}{ K _1 A} \\ R _2=\frac{2 L}{ K _2 A} \\ R _3=\frac{ L }{ K _3 A} \\ \frac{\theta-100}{\frac{ R _1 R _2}{ R _1+ R _2}}+\frac{\theta-0}{ R _3}=0 \\ \theta=40\end{array}$

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Question 44 Marks

The driver sitting inside a parked car is watching vehicles approaching from behind with the help of his side view mirror, which is a convex mirror with radius of curvature $R =2 m$. Another car approaches him from behind with a uniform speed of $90 km / hr$. When the car is at a distance of 24 m from him, the magnitude of the acceleration of the image of the side view mirror is ' a '. The value of 100a is _________ $m / s ^2$.

Answer

(8)
$
\begin{array}{l}Sol.\
v=\frac{u f}{u-f}=\frac{-24 \cdot 1}{-24-1}=\frac{24}{25} \\
m=\frac{-v}{u}=-\frac{24}{25(-24)}=\frac{1}{25} \\
\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\
v_1=-m^2 \cdot v_0=\frac{-1}{(25)^2} \cdot 25=\frac{-1}{25}
\\Diff.\end{array}
$
$
\begin{array}{l}
\frac{-1}{v^2}\left(\frac{d v}{d t}\right)+\frac{1}{u^2}\left(\frac{d u}{d t}\right)=0\left[\frac{d v}{d t}=v_1 ; \frac{d u}{d t}=v_0\right] \\
\frac{+2}{v^3} \cdot\left(v_1\right)^2-\frac{1}{v^2} \cdot a_1-\frac{2}{u^3} \cdot\left(v_0\right)^2+\frac{1}{u^2} \cdot a_0=0 \\
\frac{a_1}{v^2}=\frac{2}{v^3} \cdot v_1^2-\frac{2}{u^3} \cdot v_0^2 \\
a_1=\frac{2}{v} \cdot v_1^2-\frac{2 v^2}{u^3} \cdot v_0^2 \\
=\frac{2 \cdot 25}{24} \cdot \frac{1}{25} \cdot \frac{1}{25}-\frac{2}{(24)^3} \cdot \frac{24}{25} \cdot \frac{24}{25} \cdot 25 \cdot 25 \\
a_1=\frac{2}{24.25}-\frac{2}{24} \\
a_1=\frac{2}{24} \cdot \frac{-24}{25}=\frac{-2}{25} \\
100 a_1=\frac{2}{25} \times 100=8
\end{array}
$

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Question 54 Marks

Two soap bubbles of radius 2 cm and 4 cm , respectively, are in contact with each other. The radius of curvature of the common surface, in cm , is _________ .

Answer

(4)
Sol. $\quad r=\frac{r_1 \cdot r_2}{r_1-r_2}=\frac{2 \cdot 4}{4-2}=4 cm$

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JEE SECTION - B [PHYSICS - NUMERIC]

SECTION - B [PHYSICS - NUMERIC]

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