Physics Test

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Physics Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

The potential energy of a particle varies with distance \(x\) from a fixed origin as \(V=\left(\frac{A \sqrt{x}}{x+B}\right),\) where \(A\) and \(B\) are constants. The dimensions of \(AB\) are :

A
[ML^5/2T^-2]

B
[ML²T^-2]

C
[M^3/2L^3/2T^-2]

D
[ML^7/2T^-2]

Solution

Given, \(V =\frac{A \sqrt{x}}{x+B}\)
Dimensions of \(V =\) dimensions of potential energy \(=\left[ ML ^{2} T ^{-2}\right]\)
From Eq. (i),
Dimensions of \(B=\) dimensions of \(x=\left[ M ^{0} LT ^{0}\right]\)
Dimensions of \(A=\frac{\text { dimensions of } V \times \text { dimensions of }(x+B)}{\text { dimensions of } \sqrt{x}}\) \(=\frac{\left[ ML ^{2} T ^{-2}\right]\left[ M ^{0} L T ^{0}\right]}{\left[ M ^{0} L ^{1 / 2} T ^{0}\right]}\)
\(=\left[ ML ^{5 / 2} T ^{-2}\right]\)
Hence, dimensions of \(A B\)
\(=\left[ ML ^{5 / 2} T ^{-2}\right]\left[ M ^{0} LT ^{0}\right]\)
\(=\left[ ML ^{7 / 2} T ^{-2}\right]\)

Hence option D is the correct answer.
Question 2
1 / -0

The equation of transverse wave is given by \(y=5 \sin 2 \pi\left(\frac{t}{0.04}-\frac{x}{40}\right),\) where distance is in \(cm\) and time in second. The wavelength of wave will be :

A
20 cm

B
40 cm

C
60 cm

D
None of these

Solution

The standard equation of transverse wave is

\(y=a \sin 2 \pi\left[\frac{t}{T}-\frac{x}{\lambda}\right] \ldots \ldots(i)\)

where \(\lambda\) is wavelength and \(a\) is amplitude. Given, equation is

\(y=5 \sin 2 \pi\left[\frac{t}{0.04}-\frac{x}{40}\right] \ldots \ldots . . . .(i i)\)

Comparing Eqs. (i) and (ii), we get

\(\frac{x}{\lambda}=\frac{x}{40}\)

\(\lambda=40 \mathrm{cm}\)

Hence option B is the correct answer.
Question 3
1 / -0

A body of mass \(40 kg\) is resting on a rough horizontal surface and is subjected to a force \(P\), which is just enough to start the motion of the body. If \(\mu_{s}=5, \mu_{k}=0.4, g =10 m / s ^{2}\) and the force \(P\) is continuously applied on the body, then the acceleration of the body is :

A
zero

B
1 m/s²

C
2 m/s²

D
2.4 m/s²

Solution

since \(\mathrm{P}\) is the force which is just enough to start the motion of the body,

\(\mathrm{P}=\mu_{\mathrm{s}} \mathrm{N}=\mu_{\mathrm{s}} \mathrm{mg}\)

When the body moves under this force, the net force on body is given by \(\mathrm{F}=\mathrm{ma}=\mathrm{P}-\mathrm{f}\)

\(\Longrightarrow \mathrm{ma}=\mu_{\mathrm{s}} \mathrm{mg}-\mu_{\mathrm{k}} \mathrm{N}\)

\(=\mathrm{m}\left(\mu_{\mathrm{s}}-\mu_{\mathrm{k}}\right) \mathrm{g}\)

\(\Longrightarrow \mathrm{a}=\left(\mu_{\mathrm{s}}-\mu_{\mathrm{k}}\right) \mathrm{g}\)

\(=(0.5-0.4)(10) \mathrm{m} / \mathrm{s}^{2}\)

\(\Longrightarrow \mathrm{a}=1 \mathrm{m} / \mathrm{s}^{2}\)

Hence option B is the correct answer.
Question 4
1 / -0

At a given place where acceleration due to gravity is 'g' \(m / s^{2},\) a sphere of lead of density 'd' \(kg / m ^{3}\) is gently released in a column of liquid of density \(\rho kg / m ^{3}\). If \(d >\rho\), the sphere will :

A
fall vertically with an acceleration 'g'

B
fall vertically with no acceleration

C

fall vertically with an acceleration \(g \left(\frac{ d -\rho}{ d }\right)\)

D

fall vertically with an acceleration \(g \left(\frac{\rho}{ d }\right)\)

Solution

Apparent weight \(=\) actual weight - upthrust

\(V d g^{\prime}=V d g-V \rho g\)

\(\Rightarrow \quad g^{\prime}=\left(\frac{d-\rho}{d}\right) g\)

Hence option C is the correct answer.
Question 5
1 / -0

If the binding energies per nuclei in \({ }_{3}^{7} Li\) and \({ }_{2}^{4}\) He nuclei are \(5.60 MeV\) and \(7.06 MeV\), respectively, then in the
reaction \(p+\frac{7}{3} L i \Rightarrow 2{ }_{2}^{4}\) He \(,\) energy of proton must be :

A
28.24 MeV

B
17.28 MeV

C
1.46 MeV

D
39.2 MeV

Solution

E_p + 7 × 5.60 = 2 × 4 × 7.06

⇒ E_p = 17.28 MeV.

Hence option B is the correct answer.
Question 6
1 / -0

Two billiard balls are rolling on a flat table. One has velocity components V_x=1 m/s,V_y=√3 m/s and the other has component V_x=2 m/s and V_y=2 m/s. If both the balls start moving from the same point the angle between their path is:

A
60°

B
45°

C
22.5°

D
15°

Solution

Components of the relative velocity of the billiard balls ar \(V_{x}=1 m s^{-1}\)
\(V_{y}=(2-\sqrt{3}) m s^{-1}\)
\(\tan \theta=\frac{V_{y}}{V_{x}}=\frac{2-\sqrt{3}}{1}\)
i.e.\(\theta=\tan ^{-1} \frac{2-\sqrt{3}}{1}=15^{\circ}\)
\(\theta\) can also be found by using the formula of \(\tan (2 \theta)\).
\(\tan (2 \theta)=\frac{2 \tan (\theta)}{1-\tan ^{2}(\theta)}\)
\(\begin{aligned} &=\frac{2(2-\sqrt{3})}{1-(2-\sqrt{3})^{2}} \\ &=\frac{2(2-\sqrt{3})}{4 \sqrt{3}-6} \\ &=\frac{2(2-\sqrt{3})}{2 \sqrt{3}(2-\sqrt{3})} \\ &=\frac{1}{\sqrt{3}} \end{aligned}\)

Hence option D is the correct answer.
Question 7
1 / -0

Vector \(\overrightarrow{ Q }\) has a magnitude of 8 is added to the vector \(\overrightarrow{ P }\) which lies along the \(X\) -axis. The resultant of these two vectors is a third vector \(\overrightarrow{ R }\) which lies along the \(Y\) -axis and has a magnitude twice that of \(\overrightarrow{ P }\). The magnitude of \(\overrightarrow{ P }\) is:

A
6/√5

B
8/√5

C
12/√5

D
16/√5

Solution

Let the \(\vec{P}\) be \(x \hat{i}\). Then the \(\vec{Q}\) has to be \(-x \hat{i}+2 x j\) to make the resultant vector in \(y\) direction and double the magnitude of \(\vec{P}\).

Now, magnitude of \(\vec{P}\) is x. Using \(\sqrt{\left(x^{2}+4 x^{2}\right)}=8\) (here magnitude of \(\vec{Q}\) is 8 ) \(\Rightarrow x=\frac{8}{\sqrt{5}}\)

Hence option B is the correct answer.
Question 8
1 / -0

The distance between the point P(x,y,z) and plane xz is :

A
x

B
y

C
z

D
xy

Solution

The coordinate of any point in xz plane is \(Q(x, 0, y)\) Thus, the distance between point \(P\) and \(Q\) is \(=\sqrt{(x-x)^{2}+(y-0)^{2}+(z-z)^{2}}=y\)

Hence option B is the correct answer.
Question 9
1 / -0

The minimum number of coplanar vectors having different non-zero magnitudes can be added to give zero resultant is :

A
2

B
3

C
4

D
5

Solution

The minimum number of coplanar vectors of different magnitude to produce a resultant of zero magnitude is 3.

Hence option B is the correct answer.
Question 10
1 / -0

\begin{equation}\text { The value of } m \text { if } i +2 \overrightarrow{ j }-3 \overrightarrow{ k } \text { is parallel to } 3 \overrightarrow{ i }+ m \overrightarrow{ j }-9 \overrightarrow{ k } \text { is : }\end{equation}

A
12

B
9

C
6

D
3

Solution

If they are parallel, the components must be in same ratio. Hence \(\frac{1}{3}=\frac{2}{m}=\frac{-3}{-9}=>m=6\)

Hence option C is the correct answer.

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Physics Test - 9

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Physics Test - 9

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SHARING IS CARING

Question 1

[ML5/2T-2]

[ML2T-2]

[M3/2L3/2T-2]

[ML7/2T-2]

Solution

Question 2

20 cm

40 cm

60 cm

None of these

Solution

Question 3

zero

1 m/s2

2 m/s2

2.4 m/s2

Solution

Question 4

fall vertically with an acceleration 'g'

fall vertically with no acceleration

fall vertically with an acceleration \(g \left(\frac{ d -\rho}{ d }\right)\)

fall vertically with an acceleration \(g \left(\frac{\rho}{ d }\right)\)

Solution

Question 5

28.24 MeV

17.28 MeV

1.46 MeV

39.2 MeV

Solution

Question 6

60°

45°

22.5°

15°

Solution

Question 7

6/√5

8/√5

12/√5

16/√5

Solution

Question 8

x

y

z

xy

Solution

Question 9

2

3

4

5

Solution

Question 10

12

9

6

3

Solution

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[ML^5/2T^-2]

[ML²T^-2]

[M^3/2L^3/2T^-2]

[ML^7/2T^-2]

1 m/s²

2 m/s²

2.4 m/s²