Tag: physics

Questions Related to physics

The resultant of two equal forces acting at right angles to each other is $1414$ dyne. Find the magnitude of either force.

physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. $960$

  2. $1000$

  3. $1200$

  4. $1414$

Show answer
Correct Option: B
Explanation:

Given,

$F _1=F _2=F(say)$
$\theta=90^0$
$R=1414dyne$
Resultant force,
$R=\sqrt{F _1^2+F _2^2+2F _1F _2cos\theta}$
$1414dyne=\sqrt{F^2+F^2+2F^2cos90^0}=\sqrt{2F^2}$
$1414dyne=\sqrt{2}F$
$F=\dfrac{1414}{\sqrt{2}}=999.84 \approx 1000dyne$ (by rounding off)
The correct option is B.

A particle is moving with a velocity $35\ m/s$ along positive x-axis.It's acceleration is towards negative X-axis with the magnitude $4m/s^ {2}$ then the distance covered by the particle in the $9th$ second:-

physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. $1m$

  2. $\dfrac {9}{8}m$

  3. $\dfrac {5}{4}m$

  4. $153m$

Show answer
Correct Option: B

Two constant force $ \overrightarrow F _1 and \overrightarrow F _2 $ acts on a body.these forces displaces the body from point P(1, -2, 3) to Q (2, 3, 7 ) in 2s starting from rest.force $\overrightarrow F _1 $ is of magnitude 9 N and acting along vector $ ( 2 \hat i - 2 \hat j + \hat k ) $ . the positions are in meter. find work done by $ \overrightarrow F _1 $.

physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. $-12J$

  2. $+12J$

  3. $36J$

  4. $-36J$

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Correct Option: A

The vector sum of three forces having magnitudes $ | \overrightarrow F _1 | = 100 N $,  $ | \overrightarrow F _2 | = 80 N $ &  $ | \overrightarrow F _3 | = 60 N $ acting on a particle is zero. the angle between $ \overrightarrow F _1$  & $ \overrightarrow F _2 $ is nearly:-

physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. $ 53^0 $

  2. $ 143^0 $

  3. $ 37^0 $

  4. $ 127^0 $

Show answer
Correct Option: C
The Earth's radius is $6371km$. The order magnitude of the Earth's radius is
physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. $10^3$

  2. $10^2$

  3. $10^7$

  4. $10^5$

Show answer
Correct Option: C
Explanation:
Order of magnitude is usually written as $10 $ to the $n$th power. The represents the order of magnitude.
e.g., if we write a number $X$ in such a way that
$X = m × 10^n$ then n is order of magnitude.
here, $6371 km = 6371000 m$
$6371000 = 6.371 × 10^6$
we see, $ 6.371 ≥ \sqrt{10}$
so, $0.6371 × 10^7$
, $\dfrac{\sqrt{10}}{\sqrt{10}} ≤ 0.6371 ≤ √10$
7 is our order of magnitude.

Find the order of magnitude of the mass of the star, whose radius is $384 \times 10^6$ m and average density is $4 \times 10^3$ kg $m^{-3}$ :

physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. 30

  2. 29

  3. 24

  4. 21

Show answer
Correct Option: A
Explanation:

Radius of star $r=384\times10^6m$

So, Volume of star is $V=4\pi r^3/3=2.371\times10^{26}m^3$
Density of star is $\rho=4\times10^3kgm^{-3}$
Mass of star is $M=V\times\rho\approx9.5\times10^{29}kg=0.95\times10^{30}kg$
So the mass is of order of $10^{30}kg$

Round off following number upto four significant figures.

$45.689$

physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. $45.689$

  2. $45.69$

  3. $45.68$

  4. $45.7$

Show answer
Correct Option: B
Explanation:
The given number has $5$ significant figures.
As  $9>5$, thus the digit $8$ gets changed to  $9$ after rounding off upto four significant figures.
$\therefore$ After rounding off, $45.689$ becomes $45.69$

Round off 2.0082 to four significant figures.

physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. 2.008

  2. 2.0082

  3. 2.009

  4. 2

Show answer
Correct Option: A
Explanation:

$2.0082=2.008$, 

Here the digit $2$ is less than $5$ so $2.008$ is the closest value in four significant figures.

On a recent trip, Cindy drove her car 290 miles, rounded to the nearest 10 miles, and used 12 gallons of gasoline, rounded to the nearest gallon. The actual number of miles per gallon that Cindy's car got on this trip must have been between.

physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. $\displaystyle \frac { 290 }{ 12.5 } $ and $\displaystyle \frac { 290 }{ 11.5 } $

  2. $\displaystyle \frac { 295 }{ 12 } $ and $\displaystyle \frac { 285 }{ 11.5 } $

  3. $\displaystyle \frac { 285 }{ 12 } $ and $\displaystyle \frac { 295 }{ 12 } $

  4. $\displaystyle \frac { 285 }{ 12.5 } $ and $\displaystyle \frac { 295 }{ 11.5 } $

  5. $\displaystyle \frac { 295 }{ 12.5 } $ and $\displaystyle \frac { 285 }{ 11.5 } $

Show answer
Correct Option: D
Explanation:
Here, the rounding off of numbers is tested.
Distance rounded to nearest 10 miles = 290. 
So actual distance covered may be b/w 285 and 295. 
Gasoline used rounded to nearest gallon = 12.
So the actual gas used may be b/w 11.5 and 12.5. 
Now to get range of of miles/gallon, 
least value = the least of distance/the max of gas = 285/12.5
highest value = max of distance/least of gas = 295/11.5 
Option D is the correct choice.

The radius of the sun is $7\times 10^8 m$ and its mass is $2\times 10^{30} kg$. What is the order of magnitude of density of the sun?

physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. $1.4\times 10^3 kg/m^3$

  2. $ 10^7 kg/m^3$

  3. $1.5\times 10^3 kg/m^3$

  4. $10^3 kg/m^3$

Show answer
Correct Option: D
Explanation:

Given :  $R = 7\times 10^8 \ m$     and    $M = 2\times 10^{30} \ kg$
Volume of sun  $V = \dfrac{4}{3}\pi R^3 = \dfrac{4}{3}\pi\times (7\times 10^8)^3 = 1.4\times 10^{27} \ m^3$
Density of sun   $\rho = \dfrac{M}{V} = \dfrac{2\times 10^{30}}{1.4\times 10^{27}} = 1.43\times 10^3 \ kg/m^3$
Thus order of magnitude of density of sun is $10^3 \ kg/m^3$.