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Chapter 1 - Measurements - 9th Science Guide Samacheer Kalvi Solutions - Pdf Download [2024-2025]
Chapter 1 - Measurements - 9th Science Guide Samacheer Kalvi Solutions - Tamil Medium - Pdf Download [2024-2025]

# Chapter 1 - Measurements - 9th Science Guide Samacheer Kalvi Solutions - Tamil Medium - SaraNextGen [2024-2025]

## Chapter 1 - Measurements - 9th Science Guide Samacheer Kalvi Solutions - Chapter 1 - Measurements - 9th Science Guide Samacheer Kalvi Solutions - Tamil Medium

$?$\mathrm{m}$and Ezhilan argues that it is$9.46 \times 10<$sup$>12\mathrm{km}$. Who is right? Justify your answer. Answer: The magnitude of light year$=9.46 \times 10<$sup$>15\mathrm{m}$. So Inian gave a correct answer. Question$2 .$The main scale reading while measuring the thickness of a rubber ball using Vernier Caliper is$7 \mathrm{~cm}$and the Vernier scale coincidence is 6 . Find the radius of the ball. Answer: Given: The main scale reading$=7 \mathrm{~cm}$Vernier scale coincidence$=6$we know that least count of vernier$=0.01 \mathrm{~cm}$The radius of the ball$=M S R+V C \times L C
\begin{aligned}
& =7 \mathrm{~cm}+6 \times 0.01 \mathrm{~cm} \\
& =7 \mathrm{~cm}+0.06 \mathrm{~cm}=7.06 \mathrm{~cm}
\end{aligned}
$Question$3 .$Find the thickness of a five rupee coin with the screw gauge, if the pitch scale reading is$1 \mathrm{~mm}$and its head scale coincidence is 68 . Answer: Given Pitch scale reading$=1 \mathrm{~mm}$Head scale coincidence$=68$The thickness of a fire rupee coin$=\mathrm{PSR}+\mathrm{HSC} \times \mathrm{L} . \mathrm{C} \pm \mathrm{ZE}=1 \mathrm{~mm}+68 \times 0.01 \mathrm{~mm}
\begin{aligned}
& =1 \mathrm{~mm}+0.68 \mathrm{~mm} \\
& =1.68 \mathrm{~mm}
\end{aligned}
$ACTIVITY Question$1 .$Using Vernier caliper find the outer diameter of your pen cap. Answer: Question$2 .$Determine the thickness of a single sheet of your science textbook with the help of a Screw gauge. Answer: Pitch scale reading$=0.05 \mathrm{~mm}$L.C.$=0.1 \mathrm{~mm}$Head scale coincidence$=02$The thickness of a single sheet of science text book$=\mathrm{PSR}+\mathrm{HSC} \times \mathrm{L} . \mathrm{C} .+\mathrm{ZE}=0.05 \mathrm{~mm}+(02 \times 0.1)=0.05 \mathrm{~mm}+0.2 \mathrm{~mm}=0.07 \mathrm{~mm}$Question$3 .$With the resources such as paper plates, teacups, thread, and sticks available at home make a model of an ordinary balance. Using standard masses find the mass of some objects. Answer: -->$ ? $\mathrm{m}$ and Ezhilan argues that it is $9.46 \times 10<$ sup $>12\mathrm{km}$. Who is right? Justify your answer.
The magnitude of light year $=9.46 \times 10<$ sup $>15\mathrm{m}$. So Inian gave a correct answer.

Question $2 .$
The main scale reading while measuring the thickness of a rubber ball using Vernier Caliper is $7 \mathrm{~cm}$ and the Vernier scale coincidence is 6 . Find the radius of the ball.
Given: The main scale reading $=7 \mathrm{~cm}$
Vernier scale coincidence $=6$
we know that least count of vernier $=0.01 \mathrm{~cm}$
The radius of the ball $=M S R+V C \times L C$
\begin{aligned} & =7 \mathrm{~cm}+6 \times 0.01 \mathrm{~cm} \\ & =7 \mathrm{~cm}+0.06 \mathrm{~cm}=7.06 \mathrm{~cm} \end{aligned}

Question $3 .$
Find the thickness of a five rupee coin with the screw gauge, if the pitch scale reading is $1 \mathrm{~mm}$ and its head scale coincidence is 68 .
Given Pitch scale reading $=1 \mathrm{~mm}$
Head scale coincidence $=68$
The thickness of a fire rupee coin $=\mathrm{PSR}+\mathrm{HSC} \times \mathrm{L} . \mathrm{C} \pm \mathrm{ZE}$
$=1 \mathrm{~mm}+68 \times 0.01 \mathrm{~mm}$

\begin{aligned} & =1 \mathrm{~mm}+0.68 \mathrm{~mm} \\ & =1.68 \mathrm{~mm} \end{aligned}

ACTIVITY
Question $1 .$

Using Vernier caliper find the outer diameter of your pen cap.

Question $2 .$
Determine the thickness of a single sheet of your science textbook with the help of a Screw gauge.
Pitch scale reading $=0.05 \mathrm{~mm}$ L.C. $=0.1 \mathrm{~mm}$
Head scale coincidence $=02$
The thickness of a single sheet of science text book $=\mathrm{PSR}+\mathrm{HSC} \times \mathrm{L} . \mathrm{C} .+\mathrm{ZE}$
$=0.05 \mathrm{~mm}+(02 \times 0.1)$
$=0.05 \mathrm{~mm}+0.2 \mathrm{~mm}$
$=0.07 \mathrm{~mm}$
Question $3 .$