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Maths 2 Marks Questions

Model Paper 1 · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 7 questions.

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Question 12 Marks
A hollow spherical shell is made of a metal of density $4.5\ g\ per\ cm ^3$. If its internal and external radii are $8 \ cm$ and $9 \ cm$ respectively, find the weight of the shell.
Answer
Internal radius of the hollow spherical shell, $r = 8 \ cm$
External radius of the hollow spherical shell, $R = 9 \ cm$
Therefore, Volume of the shell $=\frac{4}{3} \pi\left(R^3-r^3\right)$
$=\frac{4}{3} \pi\left(9^3-8^3\right)$
$=\frac{4}{3} \times \frac{22}{7} \times(729-512)$
$=\frac{4 \times 22 \times 217}{21}$
$=\frac{88 \times 31}{3}$
$=\frac{2728}{3} \ cm^3$
Weight of the shell $=$ volume of the shell $\times$ density per cubic $cm$
$=\frac{2728}{3} \times 4.5 \approx 4092 g=4.092 \ kg$
Therefore Weight of the shell $=4.092 \ kg$
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Question 22 Marks
The radii of two cones are in the ratio $2 : 1$ and their volumes are equal. What is the ratio of their heights?
Answer
Radii of two cones are in the ratio of $= 2 : 1$
Let $r_1, r_2$ be the radii of two cones and $h_1, h_2$ be their respective heights .
Then $\frac{r_1}{r_2}=\frac{2}{1}$
Now, $\frac{\text { Volume of first come }}{\text { Volume of the second come}}$
$=\frac{\frac{1}{3} \pi r_1^2 h_1}{\frac{1}{3} \pi r_2^2 h_2}$
$=\frac{r_1^2 h_1}{r_2^2 h_2}$
$=\left(\frac{r_1}{r_2}\right)^2 \times\left(\frac{h_1}{h_2}\right)$
$=\left(\frac{2}{1}\right)^2 \times \frac{h_1}{h_2}$
$=\frac{4 h_1}{h_2}$
$\because$ Their volumes are equal
$\therefore \frac{4 h_1}{h_2}=1$
$\Rightarrow \frac{h_1}{h_2}=\frac{1}{4}$
$\therefore$ Their ratio is $=1: 4$
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Question 32 Marks
Prove that : $\frac{1}{3+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+1}=1$.
Answer
$\text{LHS}$
$=\frac{1}{3+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+1}$
$=\frac{1}{3+\sqrt{7}} \times \frac{3-\sqrt{7}}{3-\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}} \times \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{3}} \times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}$
$=\frac{3-\sqrt{7}}{3^2-\sqrt{7}^2}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}^2-\sqrt{5}^2}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}^2-\sqrt{3}^2}+\frac{\sqrt{3}-1}{\sqrt{3}^2-1^2}$
$=\frac{3-\sqrt{7}}{9-7}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{3}-1}{3-1}$
$=\frac{3-\sqrt{7}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{3}-1}{2}$
$=\frac{3-\sqrt{7}+\sqrt{7}-\sqrt{5}+\sqrt{5}-\sqrt{3}+\sqrt{3}-1}{2}$
$=\frac{2}{2}$
$=1$
$=\text { RHS }$
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Question 42 Marks
If $x=3+2 \sqrt{2}$, find the value of $\left(x^2+\frac{1}{x^2}\right)$.
Answer
Given, $x=3+2 \sqrt{2}$
$\therefore \frac{1}{x}=\frac{1}{(3+2 \sqrt{2})}$
$=\frac{1}{(3+2 \sqrt{2})} \times \frac{(3-2 \sqrt{2})}{(3-2 \sqrt{2})}$
$=\frac{(3-2 \sqrt{2})}{(3)^2-(2 \sqrt{2})^2}$
$=\frac{(3-2 \sqrt{2})}{(9-8)}$
$=3-2 \sqrt{2}$
$\therefore x+\frac{1}{x}=(3+2 \sqrt{2})+(3-2 \sqrt{2})$
$x+\frac{1}{x}=6$
$\Rightarrow\left(x+\frac{1}{x}\right)^2=6^2=36$
$\Rightarrow x^2+\frac{1}{x^2}+2 \times x \times \frac{1}{x}=36$
$\Rightarrow\left(x^2+\frac{1}{x^2}\right)+2=36$
$\Rightarrow\left(x^2+\frac{1}{x^2}\right)=36-2=34$
Hence, $\left(x^2+\frac{1}{x^2}\right)=34$
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Question 52 Marks
Name the quadrants in which the following points lie :
(i) p(4, 4)
(ii) Q(–4, 4)
(iii) R(–4, –4)
(iv) S(4, –4)
Answer
(i) I
(ii) II
(iii) III
(iv) IV
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Question 62 Marks
Answer
AC = BD . . . . [Given] . . . (1)
AC = AB + BC . . . . [Point B lies between A and C] . . . . (2)
BD = BC + CD . . . . [Point C lies between B and D] . . . . (3)
Substituting (2) and (3) in (1), we get
AB + BC = BC + CD
$\Rightarrow A B=C D \ldots .$ [Subtracting equals from equals]
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2 Marks Questions - Maths STD 9 Questions - Vidyadip