3 Marks Question

Question 13 Marks

In a stack, there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm . What is the total thickness of the stack?

Answer

Given, thickness of each book $=20 mm$
$\therefore$ Thickness of 5 books $=20 \times 5=100 mm$
Thickness of each paper sheet $=0.016 mm$
$\therefore$ Thickness of 5 paper sheets $=5 \times 0.016$ $=0.08 mm$
Hence, the total thickness of the stack
$=$ Thickness of 5 books
+Thickness of 5 paper sheets
$=100+0.08=100.08$
$=10008 \times 10^2$
Hence, the total thickness of the stack is $10008 \times 10^2 mm$.

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Question 23 Marks

The cells of a bacteria double in every 30 min . A scientist begins with a single cell.
(i) How many cells will be there after
(a) 10 h ? $\qquad$ (b) 25 h ?
(ii) What type of value is depicted by the cells of bacteria?

Answer

(i) The cells of a bacteria double in every 30 min i.e. number of cells of a bacteria after $30 min=2$
$\therefore$ Number of cells of a bacteria after 1 h
$
=2 \times 2=2^2=2^{2 \times 1}
$
Number of cells of a bacteria after $1 \frac{1}{2} h$
$
=2 \times 2^2=2^3=2^{2 \times \frac{3}{2}}
$
and number of cells of a bacteria after 2 h
$
=2 \times 2^3=2^4=2^{2 \times 2}
$
(a) Thus, the number of cells after 10 h
$
=2^{2 \times 10}=2^{20}
$
(b) The number of cells after 25 h
$
=2^{2 \times 25}=2^{50}
$
(ii) The value depicted by the cells of bacteria here is that it double itself after 30 min or it grows itself in $t$ h by $2^{2 \times t}$.

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Question 33 Marks

Consider a quantity of a radioactive substance. The fraction of this quantity that remains after $t$ half-lives can be found using the expression $3^{-t}$.
(i) What fraction of substance remains after 7 half-lives ?
(ii) After how many half-lives, will the fraction be $\frac{1}{243}$ of the original ?

Answer

(i) The fraction of substance remains after
$
7 \text { half-lives }=3^{-7}=\frac{1}{3^7}
$
(ii) Let after $x$ half-lives, the fraction will be $\frac{1}{243}$ of the original.
$\therefore \quad 3^{-x}=\frac{1}{243}=\frac{1}{3 \times 3 \times 3 \times 3 \times 3}=\frac{1}{3^5}$
$\begin{array}{lcl}\Rightarrow & 3^{-x}=3^{-5} \\ \Rightarrow & (3)^{-x}=(3)^{-5} & \\ \Rightarrow & -x=-5 & {[\because \text { bases are same }]} \\ \Rightarrow & x=5 & \end{array}$

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Question 43 Marks

To make ballot papers cut a sheet of paper into three. Stack the three pleces and cut the stack into three. Stack all the pieces and cut the stack into three again.

(i) Complete the table to show the number of ballot papers after five such steps :

Number of steps	Number of ballot paper
1	3
2
3
4
5

(ii) Suppose you continue this process. How many ballot papers would you have after 15 steps? How many would you have after $n$ cuts?
(iii) How many steps would it take to make atleast one lakh ballot papers?

Answer

(i)

Number of steps	Number of ballot paper
1	3
2	9
3	27
4	81
5	243

(ii) $14,348,907,3^n$
(iii) $n \geq 11$

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Question 53 Marks

Find the value of $x^{-3}$, if $x=(100)^{1-4}+(100)^0$.

Answer

Given, $x=(100)^{1-4}+(100)^0$
$x=(100)^{-3}+1 \quad\left[\because(a)^0=1\right]$
$\Rightarrow \quad x=(100)^{-3} \times \frac{1}{1}$
$\Rightarrow \quad x=\frac{1}{(100)^3} \times \frac{1}{1} \quad\left[\because a^{-m}=\frac{1}{a^m}\right]$
$\Rightarrow \quad x=\frac{1}{(100)^3}$
Now, $\quad x^{-3}=\left(\frac{1}{(100)^3}\right)^{-3}$
$\Rightarrow \quad x^{-3}=\frac{1}{\left(100^3\right)^{-3}}$
$\Rightarrow \quad x^{-3}=\frac{1}{(100)^{-9}} \Rightarrow x^{-3}=(100)^9$

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Question 63 Marks

Investigating Solar System The table shows the average distance from each planet in our solar system to the Sun.

Planet	Distance from Sun (km)	Distance from Sun in (km) Standard Notation
Earth	149600000
Jupiter	778300000
Mars	227900000
Mercury	57900000
Neptune	4497000000
Pluto	5900000000
Saturn	1427000000
Uranus	2870000000
Venus	108200000

(i) Write the distance from each planet to the Sun in standard notation.
(ii) Arrange the planets from closest to the Sun to farthest from the Sun.

Answer

(i) Earth $=1.496 \times 10^8 km$, Jupiter $=7.783 \times 10^8 km$ Mars $=2.279 \times 10^8 km$, Mercury $=5.79 \times 10^7 km$ Neptune $=4.497 \times 10^9 km$, Pluto $=5.9 \times 10^9 km$ Saturn $=1.427 \times 10^9 km$, Uranus $=2.87 \times 10^9 km$ Venus $=1.082 \times 10^8 km$
(ii) On arranging the planets from closest to the Sun to farthest from the Sun, we get following order Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, Pluto.

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Question 73 Marks

Astronomy The table shows the mass of the planets, the Sun and the Moon in our solar system.

Celestial Body	Mass (kg)	Mass (kg) in Standard Notation
Sun	1990000000000000000000000000000
Mercury	330000000000000000000000
Venus	4870000000000000000000000
Earth	5970000000000000000000000
Mars	642000000000000000000000000000
Jupiter	1900000000000000000000000000
Saturn	568000000000000000000000000
Uranus	86800000000000000000000000
Neptune	102800000000000000000000000
Pluto	12700000000000000000000
Moon	73500000000000000000000

(i) Write the mass of each celestial body in standard notation.
(ii) Arrange the planets and the Moon according to their mass, from least to greatest.
(iii) Which planet has same mass as the Earth?

Answer

(i) Mass of each celestial body in standard in notation is given below.
Mass of Sun $=1.99 \times 10^{30} kg$
Mass of Mercury $=3.3 \times 10^{23} kg$
Mass of Venus $=4.870 \times 10^{24} kg$
Mass of Earth $=5.970 \times 10^{24} kg$
Mass of Mars $=6.42 \times 10^{29} kg$
Mass of Jupiter $=1.9 \times 10^{27} kg$
Mass of Saturn $=5.68 \times 10^{26} kg$
Mass of Uranus $=8.68 \times 10^{25} kg$
Mass of Neptune $=1.02 \times 10^{26} kg$
Mass of Pluto $=1.27 \times 10^{22} kg$
Mass of Moon $=7.35 \times 10^{22} kg$
(ii) On observing the masses of the all given planets and the Moon in the exponent form, we see that the order of planets and the Moon by mass from least to greatest is
Pluto < Moon < Mercury < Venus < Earth
< Uranus < Neptune < Saturn < Jupiter < Mars
(iii) $\because$ Mass of Earth $=5.970 \times 10^{24} kg$
and mass of Venus $=4.870 \times 10^{24} kg$
Thus, we can say that planet Venus has about same mass as the Earth.

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