Physics Test

Result

Physics Test - 10

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The angles which the vector \(\vec{A}=3 \hat{i}+6 \hat{j}+2 \hat{k}\) makes with the coordinate axes are :

A
cos^-1(3/7),cos^-1(6/7)and cos^-1(2/7)

B
cos^-1(4/7),cos^-1(5/7)and cos^-1(3/7)

C
cos^-1(3/7),cos^-1(4/7)and cos^-1(1/7)

D
None of these

Solution

\(\mid f \vec{A}=A_{x} \hat{i}+A_{y} \hat{j}+A_{z} \hat{k}\)

then, the angles which it makes with coordinate axes are given by:

\[\cos \theta=\frac{\text { magnitude of component along concerned direction }}{\mathrm{A}}\]

where \(\mathrm{A}\) is the magnitude of \(\overrightarrow{\mathrm{A}}\) is \(\mathrm{A}=\sqrt{3^{2}+6^{2}+2^{2}}=7\)

therefore, angle with \(x\) - axis \(\alpha=\cos ^{-1} \frac{3}{7}\)

angle with \(y\) -axix \(\beta=\cos ^{-1} \frac{6}{7}\)

angle with z-axis \(\gamma=\cos ^{-1} \frac{2}{7}\)

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

If the vectors \(\vec{A}=2 \hat{i}+4 \hat{j}\) and \(\vec{B}=5 \vec{i}-P \hat{j}\) are parallel to each other, the magnitude of \(\vec{B}\) is :

A
5√5

B
10

C
15

D
2√5

Solution

\(\bar{A} \cdot B=|\vec{A}||\vec{B}|\)
\(10-4 P=2 \sqrt{5} \times \sqrt{25+P^{2}}\)
Squaring on both sides \(100+16 P^{2}-80 P=500+20 P^{2}\)
\(4 P^{2}+80 P+400=0\)
\(P^{2}+20 P+100=0\)
\(P=-10\)
So, magnitude of \(\bar{B}=\sqrt{25+P^{2}}=\sqrt{25+100}=\sqrt{125}=5 \sqrt{5}\)

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

The plane which can be formed with the vectors \(\bar{a}=3 \vec{i}-4 \vec{j}+2 k, \bar{b}=2 \vec{i}-\vec{j}-6 \bar{k}\)

\(\bar{c}=5 \bar{i}-5 \vec{j}-4 k\) is :

A
Quadralateral

B
Triangle

C
Circle

D
Hyperbola

Solution

Here, we observe:
\(\bar{c}=\bar{a}+\bar{b}\)
Hence, these 3 vectors can form a triangle. For eg., let x be a scalar quantity \(3 \operatorname{limes} x \bar{c}, x \bar{a}\) and \((-x) \bar{b}\) form a triangle.

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

The magnitudes of the \(x\) and \(y\) components of \(\vec{P}\) are 7 and \(6 .\) The magnitudes of the \(x\) and \(y\) components of \(\vec{P}+\vec{Q}\) are 11 and \(9,\) respectively. What is the magnitude of \(Q ?\)

A
9

B
8

C
6

D
5

Solution

Let \(\vec{R}=\vec{P}+\vec{Q}\)

\(|\vec{P}|_{x}=7,|\vec{P}|_{y}=6 \quad \& \quad|\vec{R}|_{x}=11,|\vec{R}|_{y}=9\)

\(R_{x}=P_{x}+Q_{x}\) or \(11-7=Q_{x}\) or \(Q_{x}=4\)

\(R_{y}=P_{y}+Q_{y}\) or \(9-6=Q_{y}\) or \(Q_{y}=3\)

\(Q=\sqrt{4^{2}+3^{2}}=5\)

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

Maximum and minimum magnitudes of the resultant of two vectors of magnitudes P and Q are in the ratio 3:1. Which of the following relations is true?

A
P=Q

B
P=2Q

C
PQ=1

D
none of the above

Solution

\(\frac{P+Q}{P-Q}=\frac{3}{1}\)

\(P+Q=3 P-3 Q\)

\(P=2 Q\)

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

Two forces F₁and F₂ acting at a point have a resultant F. lf F₂is doubled, F is also doubled. lf F₂ is reversed in direction, then also F is doubled. Then F₁:F₂:F= ?

A
√2:√2:√3

B
√3:√3:√2

C
√3:√2:√3

D
√3:√2:√2

Solution

\(F^{2}=F_{1}^{2}+F_{2}^{2}+2 F_{1} F_{2} \cos \theta\)
\(4 F^{2}=F_{1}^{2}+4 F_{2}^{2}+4 F_{1} F_{2} \cos \theta\) (Doubling the value of \(\left.F_{2}\right)\)
\(4 F^{2}=F_{1}^{2}+F_{2}^{2}-2 F_{1} F_{2} \cos \theta\left(\right.\) Reversing the direction of \(\left.F_{2}\right)\)
Solving the above equations, we get, \(F_{2}^{2}=F^{2}\) and \(F_{1}^{2}=\frac{3}{2} F^{2}\) Then \(F_{1}: F_{2}: F_{3}=\sqrt{3}: \sqrt{2}: \sqrt{2}\)

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

If a body starts with a velocity \(2 \hat{i}-3 \hat{j}+11 \hat{k} m s^{-1}\) and moves with an acceleration of \(10 \hat{i}+10 \hat{j}+10 \hat{k}\) ms \(^{-2}\) then its velocity after \(0.25 s\) will be:

A
(1/2)√811 ms^-1

B

\(\sqrt{\frac{811}{2}} \mathrm{ms}^{-1}\)

C

\(\sqrt{811 \mathrm{ms}^{-1}}\)

D

\(2 \sqrt{811 \mathrm{ms}^{-1}}\)

Solution

Given that:

\(\vec{u}=2 \hat{i}-3 \hat{j}+11 \hat{k} m / s\)

\(\vec{a}=10 i+10 \hat{j}+10 \hat{k} m / s^{2}\)

\(t=0.25 s\)

By using: \(v=u+a t\) \(\vec{v}=2 \hat{i}-3 \hat{j}+11 \hat{k}+(10 \hat{i}+10 \hat{j}+10 \hat{k}) 0.25\)

\(\vec{v}=2 \hat{i}-3 \hat{j}+11 \hat{k}+(2.5 \hat{i}+2.5 \hat{j}+2.5 \hat{k}) \vec{v}=4.5 \hat{i}-0.5 \hat{j}+13.5 \hat{k}\)

Therefore magnitude of velocity \(\vec{v} v=\sqrt{4.5^{2}+0.5^{2}+13.5^{2}}\) \(v=\sqrt{20.25+0.25+182.25}\)

\(v=\sqrt{202.75}\)

\(v=\frac{1}{2} \sqrt{811} \mathrm{m} / \mathrm{s}\)

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

Three forces \(F_{1}=2 \bar{i}+\bar{j}-\bar{k} N, F _{2}=2 \bar{i}+3 \overline{ j }-3 \overline{ k } N , F_{3}= a (\bar{i}+\bar{j}-\bar{k}) N\) act simultaneously on a particle. The value of \({ }^{\prime} a^{\prime}\) so that the particle may be in equilibrium is:

A
4

B
-4

C
2

D
6

Solution

For equilibrium \(F_{1}+F_{2}+F_{3}=0\)
\(F_{1}+F_{2}=4 \bar{i}+4 \bar{j}-4 \bar{k}\)
Hence, \(F_{3}\) should be \(-4(\bar{i}+\bar{j}-\bar{k})\) for \(F_{1}+F_{2}+F_{3}=0\) and hence \(a=-4\)

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

What is the relationship between the C.G.S. and S.I. units of density?

A
1 gcm⁻³=1000 kgm⁻³

B
1 gcm⁻³=1 kgm⁻³

C
1 gcm⁻³=10 kgm⁻³

D
1 gcm⁻³=10000 kgm⁻³

Solution

CGS unit of density is \(g / cm ^{3}\). Converting to SI unit \(1 g / c m^{3}=\frac{1 \times 100 \times 100 \times 100}{1000} k g / m^{3}\)
\(1 g / c m^{3}=1000 k g / m^{3}\)

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

The dimensional formula for potential energy is:

A
M²L²T⁻²

B
M¹L⁻²T⁻²

C
M¹L²T⁻²

D
M¹L²T⁻³

Solution

Potential energy =[mgh],

=[(M¹)(LT⁻²)(L)]

=[ML²T⁻²]

Hence option C is the correct answer.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Physics Test - 10

Result

Physics Test - 10

Score

-

Rank

-

SHARING IS CARING

Question 1

cos-1(3/7),cos-1(6/7)and cos-1(2/7)

cos-1(4/7),cos-1(5/7)and cos-1(3/7)

cos-1(3/7),cos-1(4/7)and cos-1(1/7)

None of these

Solution

Question 2

5√5

10

15

2√5

Solution

Question 3

Quadralateral

Triangle

Circle

Hyperbola

Solution

Question 4

9

8

6

5

Solution

Question 5

P=Q

P=2Q

PQ=1

none of the above

Solution

Question 6

√2:√2:√3

√3:√3:√2

√3:√2:√3

√3:√2:√2

Solution

Question 7

(1/2)√811 ms-1

\(\sqrt{\frac{811}{2}} \mathrm{ms}^{-1}\)

\(\sqrt{811 \mathrm{ms}^{-1}}\)

\(2 \sqrt{811 \mathrm{ms}^{-1}}\)

Solution

Question 8

4

-4

2

6

Solution

Question 9

1 gcm−3=1000 kgm−3

1 gcm−3=1 kgm−3

1 gcm−3=10 kgm−3

1 gcm−3=10000 kgm−3

Solution

Question 10

M2L2T−2

M1L−2T−2

M1L2T−2

M1L2T−3

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

cos^-1(3/7),cos^-1(6/7)and cos^-1(2/7)

cos^-1(4/7),cos^-1(5/7)and cos^-1(3/7)

cos^-1(3/7),cos^-1(4/7)and cos^-1(1/7)

(1/2)√811 ms^-1

1 gcm⁻³=1000 kgm⁻³

1 gcm⁻³=1 kgm⁻³

1 gcm⁻³=10 kgm⁻³

1 gcm⁻³=10000 kgm⁻³

M²L²T⁻²

M¹L⁻²T⁻²

M¹L²T⁻²

M¹L²T⁻³