Mathematics Test

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Mathematics Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

A pendulum bob is suspended on a flat car that moves with velocity v_o.The flat car is stopped by a bumper

(i) What is the angle through which the pendulum swings.

(ii) If the swing angle is$\theta=60^{\circ}$ and $l=10 \mathrm{m}$what was the initial speed of the flat car?

A
$2 \sin ^{-1}\left(\frac{v_{0}}{3 \sqrt{\mathrm{g}^{\prime}}}\right) 5 \mathrm{m} / \mathrm{s}$

B
$2 \sin ^{-1}\left(\frac{v_{0}}{2 \sqrt{\mathrm{g}} t}\right) 10 \mathrm{m} / \mathrm{s}$

C
$3 \sin ^{-1}\left(\frac{v_{0}}{4 \sqrt{\mathrm{g}^{l}}}\right) 15 \mathrm{m} / \mathrm{s}$

D
$4 \sin ^{-1}\left(\frac{v_{0}}{2 \sqrt{g l}}\right) 20 \mathrm{m} / \mathrm{s}$

Solution

By definition of bulk modulus,

$\begin{aligned} \mathrm{B}_{\mathrm{W}} &=-\mathrm{V} \frac{\Delta \mathrm{p}}{\Delta \mathrm{v}}=-100 \times \frac{\left(100 \times 1.013 \times 10^{5}\right)}{(99.5-100)} \\ &=2.026 \times 10^{9} \mathrm{N} / \mathrm{m}^{2} \end{aligned}$

Now as isothermal elasticity of a gas is equal to its pressure,

$\mathrm{B}_{\mathrm{A}}=\mathrm{E}_{\theta}=\mathrm{p}_{0}=1.013 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}$

So that $\frac{B u}{B_{A}}=\frac{C_{A}}{C_{W}}=\frac{2.026 \times 10^{9}}{1.013 \times 10^{5}}=2 \times 10^{4} \quad\left[\right.$ as $\left.C=\frac{1}{B}\right]$

i.e., bulk modulus of water is very large as compared to air. This means that air is about 20,000 times more compressive than water, i.e., the average distance between air molecules is much larger than between water molecules.

Hence, the correct option is (B)
Question 2
1 / -0

Two particles each of mass m are placed at A and C are such AB = BC = L. The gravitational force on the third particle placed at D at a distance L on the perpendicular bisector of the line AC is :

A
$\frac{G m^{2}}{\sqrt{2} L^{2}}$ along $\mathrm{BD}$

B
$\frac{G m^{2}}{\sqrt{2} L^{2}}$ along DB

C
$\frac{G m^{2}}{L^{2}}$ along $\mathrm{AC}$

D
none of these

Solution

since $A B=B C=D B=L,$ we have $A D=\sqrt{2} L$ Now force between particles placed at A and D is:
$F=G m m /(\sqrt{2} L)^{2}=G m^{2} / 2 L^{2}$
Similar force particle at $C$ will exert on particle at D. So total force on particle $D$ will be $2 F .$ Now the resultant component of these two forces along DB will be:
$2 \mathrm{Fcos} 45^{\circ}$
since $A B=B C=B D=L$ and $\angle D B C=\angle D B A=90^{\circ},$ we get $\angle A D B=\angle B D C=45^{\circ}$
So, the resultant force on mass $m$ at $D$ will be $\frac{G m^{2}}{\sqrt{2} L^{2}}$
Hence, the correct option is (B)
Question 3
1 / -0

A cylinder of mass $M_{c}$ and sphere of mass $M_{s}$ are placed at points A and B of two inclines, respectively. (See Figure). If they roll on the incline without slipping such that their accelerations are the same, then the ratio $\frac{\sin \theta_{c}}{\sin \theta_{s}}$ is :

A
$\sqrt{\frac{8}{7}}$

B
$\sqrt{\frac{15}{14}}$

C
$\frac{8}{7}$

D
$\frac{15}{14}$

Solution

$a=\frac{g \sin \theta}{1+\frac{k^{2}}{r^{2}}}$
Substituting $k$ for cylinder as $\frac{1}{\sqrt{2}} r$ and for sphere as $\sqrt{\frac{2}{5}} r$ we get The acceleration of a cylinder rolling down an incline is $a_{c}=\frac{2}{3} g \sin \theta_{c}$
Similarly, for sphere, $a_{s}=\frac{5}{7} g \sin \theta_{s}$
since, the two accelerations are equal, $\frac{\sin \theta_{c}}{\sin \theta_{s}}=\frac{15}{14}$
Hence, the correct option is (D)
Question 4
1 / -0

The gravitational potential of two homogeneous spherical shells of same surface density at their respective center are 3V and 4V. If the two shells coalesce into single one keeping surface density same, the potential at a point inside the shell is :

A
12V

B
7V

C
5V

D
(12/7)V

Solution

We have $M_{A}=\sigma 4 \pi R_{A}^{2}$ and $M_{B}=\sigma 4 \pi R_{B}^{2},$ where $\sigma$ is the surface density
$V_{A}=\frac{-G M_{A}}{R_{A}}, V_{B}=\frac{G M_{B}}{R_{B}}$
Or $\frac{V_{A}}{V_{B}}=\frac{M_{A} R_{B}}{M_{B} R_{A}}$
$=\frac{\sigma 4 \pi R_{A}^{2}}{\sigma 4 \pi R_{B}^{2}} \frac{R_{B}}{R_{A}}=\frac{R_{A}}{R_{B}}$
We have $V_{A}=3 V$ and $V_{B}=4 V$ Thus $\frac{V_{A}}{V_{B}}=\frac{R_{A}}{R_{B}}=3 / 4$
then $R_{B}=\frac{4}{3} R_{A}$
For new shell of mass $\mathrm{M}$ and radius $\mathrm{R}$ $M=M_{A}+M_{B}=\sigma 4 \pi R_{A}^{2}+\sigma 4 \pi R_{B}^{2}$
$\sigma 4 \pi R^{2}=\sigma 4 \pi\left(R_{A}^{2}+R_{B}^{2}\right)$
then $\frac{V}{V_{A}}=\frac{M}{A} \frac{R_{A}}{R_{B}}=\frac{\sigma 4 \pi\left(R_{A}^{2}+R_{B}^{2}\right)}{\left(R_{A}^{2}+R_{B}^{2}\right)^{1 / 2}}=\frac{R_{A}}{\sigma 4 \pi R_{A}^{2}}=\frac{\sqrt{R_{A}^{2}+R_{B}^{2}}}{R_{A}}=\frac{5}{3}$
As $V_{A}=3 V$. Thus we get the potential of the combined shell as $5 \mathrm{V}$
Hence, the correct option is (C)
Question 5
1 / -0

Two masses, 800kg and 450kg are at a distance 25m apart. The magnitude of gravitational field intensity at a point 20m distant from the 800kg mass and 15m distant from the 450kg mass is√x G (in N/kg). Then x is (where G is universal gravitational constant) :

A
6

B
4

C
7

D
8

Solution

Gravitational field due to $800 k g=\frac{800 G}{20^{2}}$
Gravitational field due to $400 \mathrm{kg}=\frac{450 \mathrm{G}}{15^{2}}$ $15^{2}+20^{2}=25^{2}$
So, the fields are also at right angles. Resultant field $=\sqrt{\left(\frac{800 G}{20^{2}}\right)^{2}+\left(\frac{450 G}{15^{2}}\right)^{2}}$
$=G \sqrt{\left(\frac{800}{400}\right)^{2}+\left(\frac{450}{225}\right)^{2}}$
$=G \sqrt{8}$
Hence, the correct option is (D)
Question 6
1 / -0

The magnitude of gravitational field at distances r₁ and r₂ from the centre of a uniform sphere of radius R and mass M is given by F1 and F2 respectively. Then which of the following is true?

(If r₁<R and r₂<R)

A
$\frac{F_{1}}{F_{2}}=\frac{r_{1}}{r_{2}}$

B
$\frac{F_{1}}{F_{2}}=\frac{r_{1}^{2}}{r_{2}^{2}}$

C
Both (a) and (b)

D
None of above

Solution

We know, $\mathrm{F}=\frac{\mathrm{GM}}{\mathrm{r}^{2}},$ if $\mathrm{r} \geq \mathrm{R}$

Force is inversely proportional to $\mathrm{r}^{2} .$ When $\$ \$ r$ lge $\mathrm{R} \$ \$$

$\mathrm{So}$

$\frac{\mathrm{F}_{1}}{\mathrm{F}_{2}}=\frac{\mathrm{r}_{2}^{2}}{\mathrm{r}_{1}^{2}} \mathrm{W}$ hen, $\mathrm{r}_{1} \geq \mathrm{R}, \mathrm{r}_{2} \geq \mathrm{R}$

again, $\mathrm{F}=\frac{\mathrm{GMr}}{\mathrm{R}^{3}}, \mathrm{W}$ hen $\mathrm{r} \geq \mathrm{R}$

Force is directly proportional to r. When

$\mathrm{r}<\mathrm{R}$

hence, $\frac{\mathrm{F}_{1}}{\mathrm{F}_{2}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ when $, \mathrm{r}_{1}<\mathrm{R}, \mathrm{r}_{2}<\mathrm{R}$

Therefore, Option $\mathrm{D}$ none of these is the correct answer.
Hence, the correct option is (A)
Question 7
1 / -0

A particle hanging from a massless spring stretches it by 2 cm at earths surface. How much will the same particle stretch the spring at a height of 2624 km from the surface of the earth? (Radius of earth = 6400 km).

A
1 cm

B
2 cm

C
3.3 cm

D
4 cm

Solution

$g=\frac{G M}{R^{2}}$ and $g_{h}=\frac{G M}{(R+h)^{2}}$
$\Rightarrow \frac{g_{h}}{g}=\left(\frac{R}{R+h}\right)^{2}=\left(\frac{6400}{6400+2624}\right)^{2}=\frac{1}{2}$
$\Rightarrow g_{h}=\frac{g}{2}$
At the surface of earth, $m g=k x$ At the height $\mathrm{h}^{\prime} \Rightarrow m g_{h}=k x^{\prime}$
$\Rightarrow \frac{g_{h}}{g}=\frac{x^{\prime}}{x} \Rightarrow x^{\prime}=\left(\frac{g_{h}}{g}\right) x=\left(\frac{1}{2}\right)(2 c m)=1 \mathrm{cm}$
Hence, the correct option is (A)
Question 8
1 / -0

A particle hanging from a spring stretches it by 1cm at earth's surface. Radius of earth is 6400km. At a place 800km above the earths surface, the same particle will stretch the spring by-

A
0.79cm

B
1.2cm

C
4cm

D
17cm

Solution

as, $m g=k x$ where, $x=1 \mathrm{cm}, k=$ spring constant for second case $m g^{\prime}=k x^{\prime}$ Hence $\frac{g}{g^{\prime}}=\frac{x}{x^{\prime}}$
$x^{\prime}=x g^{\prime} / g$
$x^{\prime}=\frac{x g R^{2}}{(R+h)^{2} g}$
Earth's radius, $R=6400$ At height, $h=800$ $x^{\prime}=\frac{1 \times\left(6400 \times 10^{3}\right)^{2}}{\left(7200 \times 10^{3}\right)^{2}}$
$x^{\prime}=0.79 \mathrm{cm}$
Hence, the correct option is (A)
Question 9
1 / -0

A satellite is projected with a velocity √1.5 times its orbital velocity just above the earth's atmosphere. The initial velocity of the satellite is parallel to the surface. The maximum distance of the satellite from the earth will be:

A
2R

B
8R

C
4R

D
3R

Solution

According to law of conservation of energy $\frac{-G M m}{R}+\frac{1}{2} m V^{2}=\frac{-G M m}{R+h}$
$=\frac{-G M m}{R}+\frac{1}{2} m\left(\sqrt{1.5} \sqrt{\frac{G M}{R}}\right)^{2}=\frac{-G M m}{R+h}$
$=\frac{-G M m}{R}+\frac{3}{4} \frac{G M m}{R}=\frac{-G M m}{R+h}$
$\Rightarrow h=3 R$
Hence, the correct option is (D)
Question 10
1 / -0

The escape velocity for a planet is ve. A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be :

A
v_e

B
v_e/√2

C
v_e /2

D
zero

Solution

$v_{e}=\left(\frac{2 G M}{R}\right)^{0.5}$
Now, if the body reached center from surface net change in energy is zero i.e. $E_{f}-E_{i}=0$ $\left(0+\left(\frac{-G M m}{R}\right)\right)-\left(\frac{m v^{2}}{2}+\frac{-3 G M m}{2 R}\right)=0 \rightarrow v=\left(\frac{G M}{R}\right)$
Taking the ratio, we get $\frac{v}{v_{e}}=\left(\frac{G M}{R}\right)^{0.5} \times\left(\frac{R}{2 G M}\right)^{0.5}$
Thus, $v=\frac{v_{e}}{2^{0.5}}$
Hence, the correct option is (B)

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Mathematics Test - 41

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Mathematics Test - 41

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Question 1

\(2 \sin ^{-1}\left(\frac{v_{0}}{3 \sqrt{\mathrm{g}^{\prime}}}\right) 5 \mathrm{m} / \mathrm{s}\)

\(2 \sin ^{-1}\left(\frac{v_{0}}{2 \sqrt{\mathrm{g}} t}\right) 10 \mathrm{m} / \mathrm{s}\)

\(3 \sin ^{-1}\left(\frac{v_{0}}{4 \sqrt{\mathrm{g}^{l}}}\right) 15 \mathrm{m} / \mathrm{s}\)

\(4 \sin ^{-1}\left(\frac{v_{0}}{2 \sqrt{g l}}\right) 20 \mathrm{m} / \mathrm{s}\)

Solution

Question 2

\(\frac{G m^{2}}{\sqrt{2} L^{2}}\) along \(\mathrm{BD}\)

\(\frac{G m^{2}}{\sqrt{2} L^{2}}\) along DB

\(\frac{G m^{2}}{L^{2}}\) along \(\mathrm{AC}\)

none of these

Solution

Question 3

87

1514

87

1514

Solution

Question 4

12V

7V

5V

(12/7)V

Solution

Question 5

6

4

7

8

Solution

Question 6

\(\frac{F_{1}}{F_{2}}=\frac{r_{1}}{r_{2}}\)

\(\frac{F_{1}}{F_{2}}=\frac{r_{1}^{2}}{r_{2}^{2}}\)

Both (a) and (b)

None of above

Solution

Question 7

1 cm

2 cm

3.3 cm

4 cm

Solution

Question 8

0.79cm

1.2cm

4cm

17cm

Solution

Question 9

2R

8R

4R

3R

Solution

Question 10

ve

ve /√2

ve /2

zero

Solution

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$\sqrt{\frac{8}{7}}$

$\sqrt{\frac{15}{14}}$

$\frac{8}{7}$

$\frac{15}{14}$

v_e

v_e/√2

v_e /2