Basic Science Test 13

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Basic Science Test 13

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Weekly Quiz Competition

Question 1
1 / -0

The mass of a lift is 2000 kg. When the tension in the supporting cable is 28000 N, then its acceleration is

A
$30 ms^{-2} downwards$

B
$4 ms^{-2} upwards$

C
$4 ms^{-2} downwards$

D
$14 ms^{-2} upwards$

Solution

$T - mg = ma$
$28000 - 2000 \times a = 2000 a$
$8000 = 2000 a$
$a = 4 m/s$
Question 2
1 / -0

The refractive index of glass is measured and values are $1.49,1.50,1.52,1.54$ and $1.48$ . Percentage error in the measurement is

A
$0.0132$ %

B
$1.32$ %

C
$0.56$ %

D
$0.156$ %

Solution

Now mean absolute error $=({y}_{1}+{y}_{2}+{y}_{3}+{y}_{4}+{ty}_{5})/5$
$(0.01+0.00+0.02+0.04+0.002)/5=0.09/5=0.018$
relative error $=$ mean absolute error/ mean value
$=0.018/1.50=0.012$
Now % error in the measurement $=100\times relative error=100\times 0.012=1.2$ %
Question 3
1 / -0

A particle $m$ starts with zero velocity along a line $y=4d$ . the
position of particles $m$ varies as $x=A \sin \omega t$ At $$\omega
t=\pi/2$$ its angular momentum with respect to the origin

A
$m a \omega d$

B
$m \omega d/A$

C
$m A d/\omega$

D
zero

Solution

$wt =\dfrac{n}{2}$
$\Rightarrow x=A \sin \dfrac{n}{2}=A$
$v=A w \cos wt=A w \cos \pi/2$
$=0$
$L=mv \times 4 d=0$
Question 4
1 / -0

The ratio of accelerations $a_{1}$ and $a_{2}$ of block of mass m as shown in figure (i) and (ii) are

A
$1 : 1$

B
$6 : 1$

C
$1 : 3$

D
$1 : 6$

Solution

From figure
$2mg - T = 2ma$
$T - mg = ma$
$-------$
$mg = 3ma$
$a_{1} = g/3$
$T - mg = ma$
$3mg - mg = ma$
$2mg = ma$
$a_{2} = 2g$
$\therefore \dfrac{a_1}{a_2} = \dfrac{g}{3} \,.\, \dfrac{1}{2g} = \dfrac{1}{6}$
Question 5
1 / -0

A body is dropped from a height of $100\ m$ . At what height the velocity of the body will be equal to half of velocity when it hits the ground.

A
$45\ m$

B
$55\ m$

C
$65\ m$

D
$75\ m$

Solution

When it hits the ground
$v^{2} = 2\ gh .... (1)$
When velocity is half
$\left (\dfrac {v}{2}\right )^{2} = 2gH$
$\dfrac {v^{2}}{H} = 2gH$
$\Rightarrow \dfrac {2gh}{4} = 2gH$
$\Rightarrow H = \dfrac {h}{4} = \dfrac {100}{4} = 25 \Rightarrow height = 75\ m$ .
Question 6
1 / -0

A perfectly elastic ball is thrown against a wall and bounces back over the head of the thrower $(T),$ as shown in the figure:
when it leaves the thrower's hand, the ball is $2\,m$ above the ground, $4\,m$ away from the wall and has initial velocity of $10\sqrt{2}\,m/s$ at an angle of $45^\circ$ to horizontal. The ball will hit the ground at an approximate horizontal distance of

A
$18\,m$ from thrower

B
$12.2\,m$ from thrower

C
$10.6\,m$ from thrower

D
$13.8\,m$ from thrower

Solution

On using netwon's second equation we have
$h = ut - \dfrac{1}{2} \times gt^2$
$-2 = 10t - \dfrac{1}{2} \times 10t^2$
$5t^2 - 10t - 2 = 0$
$t = \dfrac{10 \pm \sqrt{100 + 40}}{2}$
$= \dfrac{10 + \sqrt{140}}{2}$
$R = 10 \times t$
Question 7
1 / -0

A rocket with a lift-off mass $3.5 \times 10^{4}\,kg$ is blasted upwards with an initial acceleration of $10\,ms^{-2}.$ Then the initial thrust of the blast is

A
$3.5 \times 10^{5}\,N$

B
$7.0 \times 10^{5}\,N$

C
$14.0 \times 10^{5}\,N$

D
$1.75 \times 10^{5}\,N$

Solution

For rocket
On balancing the force we get
$mg = F - mg$
$F = m(g + g)$
$= 3.5 \times 10^4(10 + 10)$
$F = 7 \times 10^{5}\,N$
Question 8
1 / -0

A particle is dropped from a height of $20\ m$ . Speed with which the particle will hit the ground is

A
$10\ m/s$

B
$20\ m/s$

C
$30\ m/s$

D
$40\ m/s$

Solution

We know that
Velocity at ground $= \sqrt {2gh}$
$= \sqrt {2\times 10\times 20} m/s$
$= \sqrt {400} m/s$
$= 20\ m/s$ .
Question 9
1 / -0

Initial velocity of a particle moving along straight line with constant acceleration is $(20 \pm 2) \ m/s$ , if its acceleration is $a=(5 \pm 0.1) \ m/s^2$ , then velocity of particle with error, after time $t=(10 \pm 1) \ s$ , is

A
$(70 \pm 3.1) \ m/s$

B
$(70 \pm 2) \ m/s$

C
$(70 \pm 4) \ m/s$

D
$(70 \pm 8) \ m/s$
Question 10
1 / -0

Two point masses $m$ and $3m$ are placed at distance $r$ , The moment of inertia of the system about an axis passing through the centre of mass of system and perpendicular to the line joining the point masses is

A
$\frac{3}{5}mr^2$

B
$\frac{3}{4}mr^2$

C
$\frac{3}{2}mr^2$

D
$\frac{6}{7}mr^2$

Solution

Suppose the separation $r$ then the center of mass from the mass $m_1$ will be at a distance given by $r_1=\dfrac{m_2}{m_1+m_2}r$
And the center of mass from the mass, $m_2$ will be at a diatnce given by $r_2=\dfrac{m_1}{m_1+m_2}r$
So the moment of inertia will be
$I=I_1+I_2=m_1r_1^2+m_2r_2^2=\dfrac{m_1m_2^2}{(m_1+m_2)^2}r^2+\dfrac{m_2m_1^2}{(m_1+m_2)^2}r^2$
Putting $m_1=m$ and $m_2=3m$ we get $I=\dfrac{3mr^2}{4}$

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Basic Science Test 13

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Basic Science Test 13

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Question 1

30ms−2downwards 30 ms^{-2} downwards 30ms−2downwards

4ms−2upwards 4 ms^{-2} upwards 4ms−2upwards

4ms−2downwards 4 ms^{-2} downwards 4ms−2downwards

14ms−2upwards 14 ms^{-2} upwards 14ms−2upwards

Solution

Question 2

0.01320.01320.0132%

1.321.321.32%

0.560.560.56%

0.1560.1560.156%

Solution

Question 3

maωdm a \omega dmaωd

mωd/Am \omega d/Amωd/A

mAd/ωm A d/\omega mAd/ω

zero

Solution

Question 4

1:11 : 11:1

6:16 : 16:1

1:31 : 31:3

1:61 : 61:6

Solution

Question 5

45 m45\ m45 m

55 m55\ m55 m

65 m65\ m65 m

75 m75\ m75 m

Solution

Question 6

18 m18\,m18m from thrower

12.2 m12.2\,m12.2m from thrower

10.6 m10.6\,m10.6m from thrower

13.8 m13.8\,m13.8m from thrower

Solution

Question 7

3.5×105 N3.5 \times 10^{5}\,N3.5×105N

7.0×105 N7.0 \times 10^{5}\,N7.0×105N

14.0×105 N14.0 \times 10^{5}\,N14.0×105N

1.75×105 N1.75 \times 10^{5}\,N1.75×105N

Solution

Question 8

10 m/s10\ m/s10 m/s

20 m/s20\ m/s20 m/s

30 m/s30\ m/s30 m/s

40 m/s40\ m/s40 m/s

Solution

Question 9

(70±3.1) m/s(70 \pm 3.1) \ m/s(70±3.1) m/s

(70±2) m/s(70 \pm 2) \ m/s(70±2) m/s

(70±4) m/s(70 \pm 4) \ m/s(70±4) m/s

(70±8) m/s(70 \pm 8) \ m/s(70±8) m/s

Question 10

35mr2\frac{3}{5}mr^253​mr2

34mr2\frac{3}{4}mr^243​mr2

32mr2\frac{3}{2}mr^223​mr2

67mr2\frac{6}{7}mr^276​mr2

Solution

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$30 ms^{-2} downwards$

$4 ms^{-2} upwards$

$4 ms^{-2} downwards$

$14 ms^{-2} upwards$

$0.0132$ %

$1.32$ %

$0.56$ %

$0.156$ %

$m a \omega d$

$m \omega d/A$

$m A d/\omega$

$1 : 1$

$6 : 1$

$1 : 3$

$1 : 6$

$45\ m$

$55\ m$

$65\ m$

$75\ m$

$18\,m$ from thrower

$12.2\,m$ from thrower

$10.6\,m$ from thrower

$13.8\,m$ from thrower

$3.5 \times 10^{5}\,N$

$7.0 \times 10^{5}\,N$

$14.0 \times 10^{5}\,N$

$1.75 \times 10^{5}\,N$

$10\ m/s$

$20\ m/s$

$30\ m/s$

$40\ m/s$

$(70 \pm 3.1) \ m/s$

$(70 \pm 2) \ m/s$

$(70 \pm 4) \ m/s$

$(70 \pm 8) \ m/s$

$\frac{3}{5}mr^2$

$\frac{3}{4}mr^2$

$\frac{3}{2}mr^2$

$\frac{6}{7}mr^2$