Science MCQ (1 Marks)

During first hour, $\text{Speed}=\frac{60}{1}\text{km/ h}=60\text{km/ hr}$
During second hour, $\text{Speed}=\frac{70}{1}\text{km/ h}=70\text{km/ hr}$
During third hour, $\text{Speed}=\frac{20}{1}\text{km/ h}=20\text{km/ hr}$
So, the car travels different distances in equal time intervals.
This is an example of non - uniform motion.

As we know that speed $=\frac{\text{Distance}}{\text{Time}}$
$=\frac{54\times1000}{90\times60}\begin{bmatrix}1\text{km}=1000\text{m} \\1\text{min}=60\text{s}\ \ \ \ \ \ \ \end{bmatrix}=10\text{m/s}.$

$1$ kilosecond $= 1000$ seconds.

Time taken to complete $20$ oscillations $= 32s$
Time taken to complete $1$ oscillation $ = \frac{32}{20}\text{s} = 1.6\text{s}$
Time period of a pendulum is time taken by it to complete $1$ oscillation.
Time period of pendulum is $1.6$ seconds.

As we know the equation of a straight line is $Y = mX + C$
Let us assume $X$ and $Y$ are two different physical quantity and $C$ is any constant.
and the graph between them is straight line.
According to above equation,
$\text{Y}\propto\text{X}$
Hence, $X$ and $Y$ are Directly proportional.

Distance covered $= 640m$
Time taken $= 2\ min \ 40\ sec$
$= 2 × 60 + 40 = 160s$ (because $1$ minute $= 60$ seconds)
$\therefore\text{Speed}=\frac{\text{Distance}}{\text{Time}}=\frac{640}{160}=4\text{m}/\text{s}$

A person has two hourglasses, one of $30$ seconds and the other of $1$ minute.
$1$ minute $= 60$ seconds
$H.C.F.$ of $30$ seconds and $60$ seconds $=30$ seconds
He can only measure time intervals of $30, 60$, and $90$ seconds.
But he cannot measure the time interval of $50$ seconds.

We know that, in one year, there are $365$ days.
Out of the year $2005, 2006, 2007$ and $2008$, the year $2008$ is a leap year which has $366$ days.
So, total number of days $= 365 + 365 + 365 + 366 = 1461$ days.

$1$ century $= 100$ years
A century has $100$ years.

The time interval of $5$ min can be measured by both $A$ and $B$. Since, the time interval of $30$ seconds cannot be measured by clock $A$ and can be measured by clock $B$. Also, the time interval of $4$ min $10$ seconds cannot be measured by clock $A$, but can be measured by clock $B.$

The distance covered by a train would be $240\ km.$

The World's Largest Sundial in Jaipur has been telling time for almost $300$ years.

In $24$ hour clock system, $4.15 p.m.$ will be written as
$12 + 4.15 = 12hrs + 4hrs + 15$ minutes $= 16$ hours $15$ minutes.

An odometer is a device that gives us the distance covered by a car.
Distance covered by the car in the given time is $d = 57336.0 − 57321 = 15.0\ km.$

Time period of vibration is the time taken for $1$ vibration.
Given:
$10$ vibrations are completed in $1sec.$
$1$ vibration will taken only: $\frac{1}{10}=0.1\text{sec}$

The basic unit of distance is metre $(m).$
The basic unit of time is second $(s).$
$\text{Speed}=\frac{\text{Distance}}{\text{Time}}$
Therefore, the basic unit of speed is m/s.

The time is taken by a car is $12$ hours.

The total averages speed of a car is $50\ km/ h.$

Fitbit is used for measuring time nowadays.
Playstation is a gaming console, Oculus Rift is a virtual reality headset whereas $LASER$ is a type of light source.

The speed of a car is $55\ km/ h.$

Conversion of $km/ h$ into $m/ s.$
$1km = 1000m$ and $1h = 60min = 3600s$
$\Rightarrow\ \frac{1\text{km}}{1\text{h}}=\frac{1000\text{m}}{3600\text{s}}=\frac{5}{18}\text{m/s}$

Time period of a simple pendulum is the total time taken to complete one full cycle, (i.e. from $A$ to $6$ and back to $A).$

Time period of a pendulum is defined the time taken by it to complete one oscillation.
Time period of simple pendulum $\text{T}=2\pi\sqrt{\frac{1}{\text{g}}}$ where l is the length of pendulum string and g is the acceleration due to gravity.

The time period of pendulum is $0.625.$

Boojho Home to school to again home
From home to school distance is $3\ km$ time taken is
$30\text{min}=\big(\frac{30}{60}\big)\text{hour}=0.5\text{hour}$
And in returning from school to home time taken
$20\text{min}=\big(\frac{20}{60}\big)\text{hour}=0.33\text{hour}$
$\text{Average speed}=\frac{\text{Total distance}}{\text{time taken}}$
$=\frac{(3+3)}{(0.5+0.33)}\text{km/ h}$
$=7.2\text{km/ h}$

$1$ week $= 7$days
$= 7$days $\times 24hrs/ day \times 60min/ hr \times 60sec/ min$
$= 604800s$

Case I
Speed of the car $= 40 km/h$
Time taken $= 15$ min $=\frac{15}{60}=0.25\text{h}$
$\text{Speed}=\frac{\text{Distance covered}}{\text{Time taken}}$
Distance covered, d1 = Speed $\times $ Time taken $= 40 \times 0.25 = 10km$
Case II
Speed of the car = 60km/h
Time taken $= 15$ min $=\frac{15}{60}=0.25\text{h}$
$\text{Speed}=\frac{\text{Distance covered}}{\text{Time taken}}$
Distance covered, d2 = Speed \times Time taken $= 60 \times 0.25 = 15km$
Total distance covered by the car, $d = d1 + d2 = 10 + 15 = 25km$
Therefore, the total distance covered by the car is 25km.

If the particle moves with constant velocity,
$\text{Velocity v}=\frac{\triangle\text{x}}{\triangle\text{t}}=\text{constant}$
Thus for equal time intervals, the distance traveled must be equal.
The distance traveled by object in option $C$ remains the same for each interval and is equal to $\triangle\text{x}=4\text{m}-2\text{m}=2\text{m}.$

 Given, total distance $= 3 + 3 = 6\ km,$
Total time $= 30 + 20 = 50\min$
$=\frac{50}{60}\text{h}$
$\text{Average speed}=\frac{\text{Total distance}}{\text{Total time taken}}$
$=\frac{6}{\frac{50}{60}}$
$=\frac{6}{50}\times60$ $\big[\because1\text{h}=60\text{min}\big]$
$=7.2\text{km/h}$

The standard unit of length is meter $(‘m’).$

Distance $= 54\ km$
$= 54 \times 1000\ m$
Distance $= 54000\ m$
Time $= 90$ minutes
$= 90 \times 60$
Time $= 5400s$
Therefore,
Distance $= 54000\ m$
Time $= 5400s$
$\text{Speed}=\frac{\text{Distance}}{\text{Time}}$
$\text{Speed}=\frac{54000}{5400}$
Speed $= 10\ m/ s$

Speed is defined as
$\text{Speed}=\frac{\text{Distance travelled}}{\text{time interval}}$
Hence speed depends both on time and distance travelled by the body.

$\because\text{Speed(S)}=\frac{\text{Distance(D)}}{\text{Time(T)}}$

Since, speed is constant for $A$ and $B,$ so graph for $A$ and $B$ will be a straight line as given in figure.

According to $SI$ system, the standard unit of time is second.