Basic Science Test 17

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Basic Science Test 17

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Weekly Quiz Competition

Question 1
1 / -0

A clerk starts from his house with a speed of $$ 2 kmh^{-1} $$ and reaches the office 3 min late .Next day the increase speed $$ 1 kmh^{-1}$$ and reaches the office 3 min earliar .Find the distance between his house and office .

A
$$7 km$$

B
$$ 600m $$

C
$$ 700m $$

D
$$8000m $$

Solution

Ratio of speed $$=2:3$$
hence Ratio of line $$=3:2$$ (inverse of speed)
gap in time $$=6\ min$$
New $$\Rightarrow $$ gap in Ratio $$=3-2=1=6\ min$$
$$3=18\ min$$
$$2=12\ min$$
distance between o`ffice & how $$=\dfrac {12}{60}\times 3=600\ m$$
Question 2
1 / -0

A vernier callipers (with least count = 0 .1 mm ) has 20 divisions of the vernier scale .The main scale division are of

A
0 .2 mm

B
0.5 mm

C
1.0 mm

D
2.0 mm

Solution

Answer
L.C = M .S.P - IV .S .D
L.C = 0 .1 mm
0.1 mm = M.S .D - 1.8 V.S .D
M.S.D = 0.1 + 1.9
= 2 mm
Question 3
1 / -0

A body takes $$5$$ min to cool from $$80^\circ C$$ to $$70^\circ C.$$ To cool from $$80^\circ C$$ to $$60^\circ C,$$ it will take (Room temperature = $$40^\circ C$$)

A
$$5$$ min

B
$$10$$ min

C
$$12$$ min

D
$$14$$ min

Solution

Newton's law of cooling, we have
$$mc(T_1 - T_2) = k \left [ \dfrac{(T_1 + T_2)}{2} - T_0 \right ]$$ .......(i)
$$\dfrac{mc}{K} = \left [ \dfrac{(T_1 + T_2)}{2} - T_0 \right ] / \left [ \dfrac{(T_1 - T_2)}{t} \right ]$$
$$\dfrac{mc}{K} = \left [ \dfrac{80 + 70)}{2} - 40 \right ] / \left [ \dfrac{(80 - 70)}{5} \right ]$$
Substituting
$$T_{1} = 80$$ & $$T_{2} = 60$$ & $$T_{0} = 40$$
equation (i)
$$mc \dfrac{(80 - 60)}{t} = K \left [ \dfrac{(80 + 60)}{2} - 40 \right ]$$
$$20\dfrac{mc}{t} = k \times 30$$
$$t = \dfrac{20\,mc}{30\,K}$$
$$= \dfrac{20}{30} \times \dfrac{35}{2}$$
$$= \dfrac{70}{6}$$
$$t = 11.66 = 12\,min$$
Question 4
1 / -0

A hypothetical vernier scale of a travelling microscope has 40 divisions which is equal to 38 main scale divisions if each main scale division is 1.2 mm then minimum error in the measurement of length is

A
$$ 0.6 mm $$

B
$$ 1.2 mm $$

C
$$ 0.06 mm $$

D
$$ 0.1 mm $$

Solution

We know that
$$ 40 VSR = 3 SMS R $$
$$ WSR = \frac { 38}{40} $$
least count $$ = 1 - \frac {38}{40} MSR $$
$$ = \frac {2}{40} = \frac {1}{20} \times 1.2 = 0.06 $$
Question 5
1 / -0

Glucose contains

A
$$20\%$$ carbon by mass

B
$$40\%$$ carbon by mass

C
$$50\%$$ carbon by mass

D
$$60\%$$ carbon by mass

Solution

$$M.W_{C_6 H_{2}O_6}=12\times 6+12\times 1+6\times 16$$ $$=180$$
total mass of carbon $$=12\times 6=72$$
$$\therefore \dfrac {72}{180}\times 100=40\%$$
Question 6
1 / -0

M gram of ice at $$0^{\circ} C$$ is mixed with M gram of water at $$10^{\circ} C.$$ The final temperature is

A
$$8^{\circ} C$$

B
$$6^{\circ} C$$

C
$$4^{\circ} C$$

D
$$0^{\circ} C$$

Solution

Let final temperature = T then
Heat required to melt ice and to increase temperature by T
$$H_1 = ML + MC\Delta T$$
$$T = \dfrac{-530}{2} = -265^{\circ} C$$
But it can reach only $$0^{\circ} C.$$
So partly the ice will melt and all the water will be at $$0^{\circ} C.$$
Question 7
1 / -0

Which of the following instrument is the least precise device for measring length?

A
Vernier callipers with 50 division of Vernier scale equal to 49 division of main scale and each division of main scale is of 1 mn.

B
A screw scale having 100 divisions and each division is of 1 cm.

C
A metre scale having 100 divisions and each division is of 1 cm.

D
An optical instrument that can resolve upto $$10^{-7} m$$.

Solution

Most precise device is the one with the minimum least count
1. Vernier Calliper
20 divisions on the sliding scale
L.C= $$\frac{1}{20}mm$$ = 0.05mm
2. Screw Gauge
pitch 1 mm and 100 divisions on the circular scale
L.C= $$\frac{1}{100}mm$$ = 0.01mm
3. meter Scale
pitch 1 mm and 100 divisions on the scale
L.C= $$\frac{1}{100}mm$$ = $$10^{-2}mm$$
4.optical instrument
within a wavelength of light
Wavelenght of light $$\approx$$ $589 n m$
$Hence option D is correct$
Question 8
1 / -0

Which of the statement are correct ?
(1) An electron $$4p$$ orbital has more energy than one in $$3p$$ orbital
(2) An electron in $$4s$$ orbital is further from the nucleus than one in a $$3s$$ orbital
(3) the angular quantum number $$1$$ is related to orbital shape.
(4) Energy is released if electron moves from $$n=2$$ to $$n=5$$ level.

A
$$1,2$$ and $$3$$

B
$$1$$ and $$3$$

C
$$2$$ and $$4$$

D
Only $$4$$

Solution

$$n+l$$rule
1) $$4p=4+1$$
$$3p=3+1$$
$$4p>3p$$
2) energy of $$4s>3s$$ so $$4s$$ is further
3) correct
4) $$2$$ to $$5$$ energy is absorbed $$x$$
Question 9
1 / -0

A metal tape gives correct measurement at $$15^{\circ}C$$. It is used to measure a distance of $$100m$$ at $$45^{\circ}C$$. The error in the measurement, if $$\alpha = 12\times 10^{-6}/1^{\circ}C$$ is

A
$$36\ cm$$

B
$$36\ m$$

C
$$42\ mm$$

D
$$36\ mm$$

Solution

Correct measurement at $$15^{\circ}C$$
$$\% error = \dfrac {\triangle l}{l}\times 100$$
$$= \dfrac {l\alpha \triangle T}{l} \times 100$$
$$\dfrac {\triangle l}{l} \times 100 = 12\times 10^{-6} \times (45 - 15) \times 100$$
$$\triangle l = l\times 12\times 10^{-6} \times 30$$
$$= 100\times 12\times 10^{-6} \times 30$$
$$= 360\times 10^{-6} = 36\ mm$$.
Question 10
1 / -0

Which of the following temperature will read the same value on celsius and Fahrenheit scales.

A
$$-40^{\circ}$$

B
$$+40^{\circ}$$

C
$$-80^{\circ}$$

D
$$-20^{\circ}$$

Solution

We know that
$$\dfrac {C - O}{100} = \dfrac {F - 32}{180}$$ (When both temperature are equal)
$$\dfrac {C}{100} = \dfrac {C - 32}{180}$$
$$180C = 100C - 3200$$
$$180C = 100C = -3200$$
$$80C = -3200$$
$$\therefore C = \dfrac {-3200}{80} = -40^{\circ}C$$
$$F = -40^{\circ}F$$.

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Basic Science Test 17

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Basic Science Test 17

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Question 1

$$7 km$$

$$ 600m $$

$$ 700m $$

$$8000m $$

Solution

Question 2

0 .2 mm

0.5 mm

1.0 mm

2.0 mm

Solution

Question 3

$$5$$ min

$$10$$ min

$$12$$ min

$$14$$ min

Solution

Question 4

$$ 0.6 mm $$

$$ 1.2 mm $$

$$ 0.06 mm $$

$$ 0.1 mm $$

Solution

Question 5

$$20\%$$ carbon by mass

$$40\%$$ carbon by mass

$$50\%$$ carbon by mass

$$60\%$$ carbon by mass

Solution

Question 6

$$8^{\circ} C$$

$$6^{\circ} C$$

$$4^{\circ} C$$

$$0^{\circ} C$$

Solution

Question 7

Vernier callipers with 50 division of Vernier scale equal to 49 division of main scale and each division of main scale is of 1 mn.

A screw scale having 100 divisions and each division is of 1 cm.

A metre scale having 100 divisions and each division is of 1 cm.

An optical instrument that can resolve upto $$10^{-7} m$$.

Solution

Most precise device is the one with the minimum least count

1. Vernier Calliper20 divisions on the sliding scale

2. Screw Gaugepitch 1 mm and 100 divisions on the circular scale

3. meter Scalepitch 1 mm and 100 divisions on the scale

4.optical instrumentwithin a wavelength of lightWavelenght of light $$\approx$$ 589nm

Question 8

$$1,2$$ and $$3$$

$$1$$ and $$3$$

$$2$$ and $$4$$

Only $$4$$

Solution

Question 9

$$36\ cm$$

$$36\ m$$

$$42\ mm$$

$$36\ mm$$

Solution

Question 10

$$-40^{\circ}$$

$$+40^{\circ}$$

$$-80^{\circ}$$

$$-20^{\circ}$$

Solution

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1. Vernier Calliper
20 divisions on the sliding scale

2. Screw Gauge
pitch 1 mm and 100 divisions on the circular scale

3. meter Scale
pitch 1 mm and 100 divisions on the scale

4.optical instrument
within a wavelength of light
Wavelenght of light $$\approx$$ $589 n m$