Basic Science Test 19

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Basic Science Test 19

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Weekly Quiz Competition

Question 1
1 / -0

Moment of inertia of a thin circular plate of mass M, radius R about an axis passing through its diameter I. The moment of inertia of a circular ring of mass M, radius R about an axis perpendicular to its plane and passing through its centre is

A
$$2I$$

B
$$\dfrac{I}{2}$$

C
$$4I$$

D
$$\dfrac{I}{4}$$

Solution

We know a moment of inertia of a disc, axis passing through its diameter is $$I = \dfrac{MR^2}{4}$$
$$\therefore I' = MR^2$$
$$I'=4I$$
Question 2
1 / -0

A man and woman are both heterozygous for the recessive allele that causes cystic fibrosis. What is the probability that their first two off spring will have the disorder ?

A
$$1$$

B
$$1 / 4$$

C
$$1 / 16$$

D
$$1 / 32$$
Solution
Correct option: C

Explanation:
Cystic fibrosis is an autosomal recessive disorder, which means the cell should have both affected chromosome to become affected.
The Man and Female are both heterozygous with affected recessive allele i.e., $$X^CY$$ and $$X^CX$$ are crossed.
$$X^C$$ $$Y$$
$$X^C$$ $$X^CX^C$$ $$X^CY$$
$$X$$ $$X^CX$$ $$XY$$
Out of 4 only 1/4 is affected. Therefore, the probability of having the disorder in first two off spring would be 1/4 X 1/4 = 1/16.
Thus, Option C is correct.
Question 3
1 / -0

$$I$$ is moment of inertia of a thin circular ring about an axis perpendicular to the plane of ring and passing through its centre. The same ring is folded into 2 turns coil. The moment of inertia of circular coil about an axis perpendicular to the plane of coil and passing through its centre is

A
2I

B
4I

C
$$\frac{I} {2}$$

D
$$\frac{I} {4}$$
Question 4
1 / -0

In a new system of units, the unit of mass equals $$\alpha$$ kg, the unit of length equals $$\beta$$ m and unit of time equals $$\gamma$$ second. The value of 1 newton in this new system of units is

A
$$\dfrac{\alpha\beta}{\gamma^2}$$

B
$$\dfrac{\alpha}{\beta\gamma^2}$$

C
$$\dfrac{\gamma}{\alpha\beta^2}$$

D
$$\dfrac{\gamma^2}{\alpha\beta}$$
Question 5
1 / -0

A man holds a ball of (1/2) kg in his hand. He throws it vertically upward. During this process his hand moves up by 40 cm and the ball leaves his hand with an upward velocity of 4 m/s. The constant force with which the man pushes the ball is

A
2 N

B
10 N

C
15 N

D
7 N
Question 6
1 / -0

Let us consider a system of units in which mass and angular momentum are dimensionless. If the length has dimension of L, then the dimension of power is

A
$$L^{-2}$$

B
$$L^{-4}$$

C
$$L^{-6}$$

D
$$L^{3}$$

Solution

$$[M] = mass = M^0L^0T^0$$
$$J(R)$$ = angular momentum
$$e = mvR = \dfrac{mL^2}{T}$$
$$m \rightarrow$$ Dimensionless
$$T = L^{2}$$
L is also Dimensionless
Now $$P = mv = \dfrac{mL}{T}$$
m is dimensionless
$$[P] = L^{-1}$$
Power = $$\dfrac{mL^2}{T^2 . T} = \dfrac{L^2}{L^2T^2L} = L^{-4}$$
Question 7
1 / -0

A long capillary tube with both ends open is filled with water and set in a vertical position. The radius of the capillary is $$0.1\ cm$$ and surface tension of water is $$70$$ dynes/ cm. The length of the water column remaining in the capillary tube will be approximately.

A
$$1.5\ cm$$

B
$$2.9\ cm$$

C
$$3.6\ cm$$

D
$$5.8\ cm$$

Solution

Weight of fliuid = Surface tension force at both end
$$\pi r^2 d_0gh =2F=4\pi r\alpha$$

$$h=\dfrac {4\alpha}{d_0 gr}$$

here $$\alpha$$ is surface position

$$h=3\ cm$$

$$3.14\times (0.1)^2 \times 10^{-4}\times 100\times 10\times h$$

$$=4\times 3.14\times 0.1\times 10^{-2}x$$

$$h=\dfrac {4\times 70\times 10^{-5}/ 10^{-2}}{1000\times 10\times 0.1\times 10^{-2}}$$

$$=\dfrac {280\times 10^{-5}}{10^{-1}}$$

$$=280\times 10^{-4}m$$

$$=0.28\times 10^{-2}m$$

$$=0.28\ cm$$
Question 8
1 / -0

A torque of magnetude 200 Nm acting body produces an angular acceleration of 8 $$ rad / s ^{ 2} .$$ . The moment of inertia of the body is

A
$$ 25 kg - m ^{2} $$

B
$$ 12.5 kg - m ^{2}$$

C
$$ 50 kg - m ^{2}$$

D
$$ 75 kg - m ^{2} $$
Question 9
1 / -0

If $$a_1 $$ and $$a_2$$ are two non collinear unit vector unit and if $$|a_1 +a_2| =\sqrt 3$$, then the value of $$(a_1-a_2), (2a_1 +a_2)$$ is

A
$$2$$

B
$$\dfrac 32$$

C
$$\dfrac 12$$

D
$$113$$ There are $$N$$ coplaner vectors each of

Solution

$$|a_1 +a_2|=\sqrt 3$$
$$\Rightarrow \sqrt {1+2\cos \theta}=\sqrt 3$$
$$\Rightarrow 2+2 c \theta =3$$
$$\Rightarrow 2x \theta =1\Rightarrow c\theta =\dfrac 12$$
$$(a_1 -a_2).(2a_1 +a_2)$$
$$=2a_1^2 +a_1.a_2 -2a_1.a_2-a_2^2$$
$$(\because a_1^2 =a_2^2 =a_1 .a_1 =|a_1| (a_1) \cos \theta =1)$$
$$=2-1+\dfrac {1}{2}-\dfrac {2}{2}=\dfrac {1}{2}$$
Question 10
1 / -0

The velocity at the maximum height of a projectile is half of its initial velocity of projection. The angle of projection is

A
$$30^{\circ}$$

B
$$45^{\circ}$$

C
$$60^{\circ}$$

D
$$76^{\circ}$$

Solution

Let initially velocity is $$V$$
So, at the highest point only horizontal velocity exists i.e. $$V_{horizontal} = \dfrac{V}{2}$$
Now, $$V_{horizontal} = \dfrac{V}{2}= Vcos \theta$$
$$\implies \cos\theta = \dfrac{1}{2}$$
$$\implies \theta = 60^{\circ}$$

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	$$X^C$$	$$Y$$
$$X^C$$	$$X^CX^C$$	$$X^CY$$
$$X$$	$$X^CX$$	$$XY$$

Basic Science Test 19

Result

Basic Science Test 19

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SHARING IS CARING

Question 1

$$2I$$

$$\dfrac{I}{2}$$

$$4I$$

$$\dfrac{I}{4}$$

Solution

Question 2

$$1$$

$$1 / 4$$

$$1 / 16$$

$$1 / 32$$

Solution

Question 3

2I

4I

$$\frac{I} {2}$$

$$\frac{I} {4}$$

Question 4

$$\dfrac{\alpha\beta}{\gamma^2}$$

$$\dfrac{\alpha}{\beta\gamma^2}$$

$$\dfrac{\gamma}{\alpha\beta^2}$$

$$\dfrac{\gamma^2}{\alpha\beta}$$

Question 5

2 N

10 N

15 N

7 N

Question 6

$$L^{-2}$$

$$L^{-4}$$

$$L^{-6}$$

$$L^{3}$$

Solution

Question 7

$$1.5\ cm$$

$$2.9\ cm$$

$$3.6\ cm$$

$$5.8\ cm$$

Solution

Question 8

$$ 25 kg - m ^{2} $$

$$ 12.5 kg - m ^{2}$$

$$ 50 kg - m ^{2}$$

$$ 75 kg - m ^{2} $$

Question 9

$$2$$

$$\dfrac 32$$

$$\dfrac 12$$

$$113$$ There are $$N$$ coplaner vectors each of

Solution

Question 10

$$30^{\circ}$$

$$45^{\circ}$$

$$60^{\circ}$$

$$76^{\circ}$$

Solution

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