Properties of Matter

Practise Properties of Matter for FBISE Grade 9–10 Physics (Grade 9–10): 8 multiple-choice questions with instant checking and worked explanations. Free, no sign-up.

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1. Question 1

   Which principle states that pressure applied to an enclosed liquid is transmitted equally in all directions?
   • APascal's law
   • BBoyle's law
   • CArchimedes' principle
   • DHooke's law
   Show answer and explanation

   Answer: A) Pascal's law

   Pascal's law describes how pressure applied to a confined liquid is transmitted equally in all directions throughout the fluid. This matches the wording of the question, so the correct answer is "Pascal's law". Hooke's law deals with elastic deformation, Archimedes' principle with buoyant force, and Boyle's law with the pressure‑volume relationship of gases; none describe pressure transmission in liquids.

2. Question 2

   What is the SI unit of Young's modulus?
   • A$\mathrm{N}\mathrm{m}$
   • B$\mathrm{N}{\mathrm{m}}^{-2}$
   • C$\mathrm{N}$
   • D$\mathrm{N}{\mathrm{m}}^{-1}$
   Show answer and explanation

   Answer: B) $\mathrm{N}{\mathrm{m}}^{-2}$

   Young's modulus is the ratio of stress (force per unit area) to strain (dimensionless). Because stress is measured in newtons per square metre ($\mathrm{N}{\mathrm{m}}^{-2}$), the modulus has the same unit. Therefore the correct answer is $\mathrm{N}{\mathrm{m}}^{-2}$. The option $\mathrm{N}{\mathrm{m}}^{-1}$ is the unit of a spring constant, $\mathrm{N}\mathrm{m}$ is the unit of torque or work, and $\mathrm{N}$ is simply a force unit; none of these represent stress per unit strain.

3. Question 3

   A solid wooden block is repositioned on a table so that it rests on its smallest face instead of its largest face. How do the downward force and pressure exerted on the table change?
   • AForce remains constant and pressure increases.
   • BForce remains constant and pressure decreases.
   • CForce increases and pressure remains constant.
   • DBoth force and pressure increase.
   Show answer and explanation

   Answer: A) Force remains constant and pressure increases.

   The block’s weight (downward force) is unchanged by its orientation, so the force on the table remains constant. Pressure is defined as force divided by area; when the block rests on its smallest face the contact area decreases, making the pressure larger. The option “Force remains constant and pressure decreases” reverses the area‑pressure relationship, while the other two options incorrectly change the force.

4. Question 4

   A hydrometer is placed sequentially into two different liquids, $X$ and $Y$, and sinks deeper in liquid $X$ than in liquid $Y$. Which statement correctly compares the two liquids?
   • ALiquid $Y$ exerts a greater upthrust than liquid $X$.
   • BLiquid $Y$ is denser than liquid $X$.
   • CLiquid $X$ exerts a greater upthrust than liquid $Y$.
   • DLiquid $X$ is denser than liquid $Y$.
   Show answer and explanation

   Answer: B) Liquid $Y$ is denser than liquid $X$.

   A hydrometer reaches equilibrium when the buoyant force equals its weight. In a denser liquid a smaller displaced volume suffices, so it floats higher. Because it sinks deeper in X, X must be less dense than Y, making Y the denser liquid. The option “Liquid X is denser than liquid Y” reverses this relationship, while the two upthrust statements ignore that upthrust equals the weight in both cases.

5. Question 5

   A metal cube of volume $1.0\times 10^{-3}{\mathrm{m}}^3$ and mass $1.4\mathrm{kg}$ is completely submerged in an oil of density $800\mathrm{kg}{\mathrm{m}}^{-3}$. It is held in equilibrium by a vertical wire of cross-sectional area $1.0\times 10^{-6}{\mathrm{m}}^2$ and Young's modulus $2.0\times 10^{11}\mathrm{N}{\mathrm{m}}^{-2}$. Taking $g=10\mathrm{m}{\mathrm{s}}^{-2}$, what is the tensile strain produced in the wire?
   • A$1.1\times 10^{-4}$
   • B$4.0\times 10^{-5}$
   • C$3.0\times 10^{-5}$
   • D$7.0\times 10^{-5}$
   Show answer and explanation

   Answer: C) $3.0\times 10^{-5}$

   The weight of the metal cube is $W=mg=1.4\mathrm{kg}\times 10\mathrm{m}{\mathrm{s}}^{-2}=14\mathrm{N}$. The buoyant upthrust from the oil is $U=\rho Vg=800\mathrm{kg}{\mathrm{m}}^{-3}\times 1.0\times 10^{-3}{\mathrm{m}}^3\times 10\mathrm{m}{\mathrm{s}}^{-2}=8\mathrm{N}$. For equilibrium, the tension force $T$ in the wire is $T=W-U=14\mathrm{N}-8\mathrm{N}=6\mathrm{N}$. Using Young's modulus $E=\frac{\text{stress}}{\text{strain}}=\frac{T/A}{\text{strain}}$, the strain is $\text{strain}=\frac{T}{AE}=\frac{6}{1.0\times 10^{-6}\times 2.0\times 10^{11}}=3.0\times 10^{-5}$.

6. Question 6

   A hydraulic press has an input piston of cross-sectional area $A$ and an output piston of area $8A$. A constant downward force of $50\mathrm{N}$ is applied to the input piston. If a frictional force of $40\mathrm{N}$ opposes the motion of the output piston, what mass can be lifted at constant speed? ($g=10\mathrm{m}{\mathrm{s}}^{-2}$)
   • A$360\mathrm{kg}$
   • B$36\mathrm{kg}$
   • C$40\mathrm{kg}$
   • D$44\mathrm{kg}$
   Show answer and explanation

   Answer: B) $36\mathrm{kg}$

   According to Pascal's principle, pressure transmitted in the hydraulic fluid is $P=\frac{F_{in}}{A}=\frac{50}{A}$. The ideal upward force exerted on the output piston is $F_{out}=P\times 8A=400\mathrm{N}$. Since friction ($40\mathrm{N}$) opposes upward motion, the net force available to lift the load is $400\mathrm{N}-40\mathrm{N}=360\mathrm{N}$. At constant speed, the load's weight equals this force ($W=360\mathrm{N}$), which corresponds to a mass of $m=\frac{W}{g}=\frac{360}{10}=36\mathrm{kg}$.

7. Question 7

   A wire of original length $L$ and diameter $d$ extends by an amount $x$ under a tensile load $W$. If the wire's diameter and the applied load are both doubled, what will the new extension be?
   • A$\frac{x}{4}$
   • B$\frac{x}{2}$
   • C$x$
   • D$2x$
   Show answer and explanation

   Answer: B) $\frac{x}{2}$

   Extension $x$ is given by $x=\frac{WL}{AE}=\frac{WL}{\frac{\pi d^2}{4}E}\propto \frac{W}{d^2}$. When the diameter is doubled ($d^{\prime }=2d$), the cross-sectional area increases by a factor of $2^2=4$. Doubling the load ($W^{\prime }=2W$) doubles the extension, but quadrupling the area reduces it by a factor of $4$. Therefore, the new extension is $x^{\prime }=x\times \frac{2}{4}=\frac{x}{2}$.

8. Question 8

   A solid block of uniform density floats in a liquid such that two-fifths of its total volume remains above the surface. What is the ratio of the density of the liquid to the density of the block?
   • A$3 : 5$
   • B$2 : 5$
   • C$5 : 2$
   • D$5 : 3$
   Show answer and explanation

   Answer: D) $5 : 3$

   Because three‑fifths of the block’s volume is submerged, the liquid must be 5/3 times as dense as the block for the buoyant force to balance the weight. This gives the ratio liquid : block = 5 : 3. The option “$3 : 5$” inverts the required ratio, “$5 : 2$” uses the unsubmerged fraction instead of the submerged one, and “$2 : 5$” incorrectly relates the above‑surface fraction directly to the density ratio.

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