Properties Of Matter Practice Questions

Hot water flows faster because of less viscosity $(✔)$ and soap water has less surface tension because bubbles are easily formed ( $X$ )

$\;-\;\frac{\Delta v B}{v}=\Delta P \Rightarrow 4 \times 10^5 {\text{Pa}}=\;{2 \times 10^9 {\text{Pa}} \times 10^{-6} \times 0.8}/{v}$ $\;v=\;\frac{2 \times 10^9 \times 10^{-6} \times 0.8}{4 \times 10^5}=4 \times 10^{-3} {\text{m}}^3$ $\;\Rightarrow 4 \;\text{ litre }\;$

$\;h=\;\frac{2 T \cos \theta }{\rho g r}$ $\;=\;\frac{100}{7}$ $\;h^{'}=\;\frac{h}{\cos \theta }=\;\frac{100}{7 \cos 30^{\circ}}$ $\;h^{'}=\;{200}/{7 {\sqrt{3}}} \approx 16.49$

$\;I=I_0+\Delta I=I_0+\;\frac{T I_0}{A Y}$ $\;I=I_0(1+K T)$ $\;1.4=I_0(1+5 K)$ $\;1.56=I_0(1+7 K)$On solving$l_0=1$

The shear modulus $G$ is given by $G = \frac{\tau}{\theta} = \frac{F}{A \theta}$, where $A$ is the area of the face on which the force $F$ is applied. For a slab of side $l$ and thickness $d$, the area of the narrow face is $A = l \cdot d$.
Thus, the ratio of the shear moduli is:
$\frac{G_2}{G_1} = \frac{F / (l d_2 \theta_2)}{F / (l d_1 \theta_1)} = \frac{d_1 \theta_1}{d_2 \theta_2}$
Substituting $d_2 = 2 d_1$ and $\theta_2 = 2 \theta_1$ into the equation:
$\frac{G_2}{G_1} = \frac{d_1 \theta_1}{(2 d_1)(2 \theta_1)} = \frac{1}{4}$
Given $G_1 = 4 \times 10^9 \, \text{N/m}^2$, we calculate $G_2$:
$G_2 = \frac{G_1}{4} = \frac{4 \times 10^9}{4} = 1 \times 10^9 \, \text{N/m}^2$
Comparing this to $x \times 10^9 \, \text{N/m}^2$, we get $x = 1$.

[CORRECT_OPTION: 1]

$\;A_1 V_1=A_2 V_2$ $\;V_2= (\;\frac{A_1}{A_2}) V_1=(4) \times 2$ $\;\; =8 {\text{m}} / {\ s}$

$\;B=-\;\frac{V P}{\Delta V}$ $\;\Rightarrow \Delta V=-V \;\frac{\Delta P}{B}$ $\;=-(10 cm)^3 \times \;\frac{7 \times 10^6 Pa}{1.4 \times 10^{11} N / m ^2}=50 mm^3$

$\;2 \times \;\frac{4}{3} \pi r^3=\;\frac{4}{3} \pi R^3$ $\;R=(2)^{1 / 3} r$ $\Delta Q\;=E_i-E_f$ $\;=2 \times 4 \pi r^2 T-4 \pi (2)^{2 / 3} r^2 T$ $=\;4 \pi R^2 T (2-(2)^{2 / 3})$

The meniscus shape in a capillary tube depends on the contact angle $\theta$: it is concave if $0^\circ \le \theta < 90^\circ$ ($\cos \theta > 0$) and convex if $90^\circ < \theta \le 180^\circ$ ($\cos \theta < 0$).
Given $K = \frac{\cos \theta_A}{\cos \theta_B} < 0$, the values of $\cos \theta_A$ and $\cos \theta_B$ must have opposite signs.
This implies one liquid has a positive cosine value, resulting in a concave meniscus, while the other has a negative cosine value, resulting in a convex meniscus.
Specifically, if $\cos \theta_A > 0$ and $\cos \theta_B < 0$, liquid $A$ exhibits a concave meniscus and liquid $B$ exhibits a convex meniscus.
This satisfies the condition given in option A.

Volume of cube inside water$\;= (\;\frac{\;\text{ Density of cube }\;}{\;\text{ Density of water }\;}) \times \;\text{ Volume of cube }\;$ $\;=\;\frac{\;\text{ Mass of cube }\;}{\;\text{ Density of water }\;}$ $\;=\;\frac{400 gm}{1 gm / cm^3}$ $\;=400 cm^3$Volume of cube outside water$\;=\;\text{ Volume of cube }\;- \;\text{ Volume of cube inside water }\;$ $\;=1000 cm^{3-}400 cm^3$ $\;=600 cm^3$

$\;\text{ Strain }\;=\;\frac{\Delta L}{L}=\;\frac{\;\text{ Stress }\;}{y}=\;\frac{F}{A y}$So, $\;\frac{\Delta L(A)}{\Delta L(B)}=\;\frac{L_A A_{(B)}}{A_{(A)} L_B}=\;\frac{1}{3} \times \;\frac{1}{4}=\;\frac{1}{12}$

$\;B=\;\frac{\Delta P}{- (\;\frac{\Delta V}{V}) }$ $\;- (\;\frac{\Delta V}{V}) =\;\frac{\Delta P}{B}=\;\frac{\rho g h}{B}$ $\; \;\; \;\; =\;\frac{10^3 \times 10 \times 2.5 \times 10^3}{2 \times 10^9}=1.25 \%$

$\;25 \times 0.2 \times g=2 \times (m-\rho \times v) g$ $\;m-\rho v=2.5 {\text{kg}}$ $\;\rho v=\;{1 \times 10^{-3} {\text{kg}}}/{ {\text{cm}}^3} \times \;{10^3 {\text{cm}}^3}/{2}=\;\frac{1}{2} {\text{kg}}$ $\;m=3 {\text{kg}}$

At terminal velocity $a=0$ $\;F_B=\;\frac{v \rho }{2} g=\;\frac{M g}{2}$ $\; \therefore \;\; F_d=M g-\;\frac{M g}{2}=\;\frac{M g}{2}$

Modulus of Elasticity (Young's Modulus, E)- Dimensional formula:$E=\;\frac{\;\text{ Stress }\;}{\;\text{ Strain }\;}$- Stress $=$ Force $/$ Area $=\;\frac{M L T^{-2}}{L^2}=M L^{-1} T^{-2}$- Strain is dimensionless.- So, dimensional formula of $E$ :$[E]=M L^{-1} T^{-2}$Torque ( $\tau$ )- Torque $=$ Force × Distance- $\tau = (M L T^{-2}) \times L=M L^2 T^{-2}$- So, dimensional formula of $\tau$ :${[\tau ]=M L^2 T^{-2}}$ $\;\frac{E}{\tau }=\;\frac{M L^{-1} T^{-2}}{M L^2 T^{-2}}$Canceling $M$ and $T^{-2}$ :Thus, the dimensional formula of the required quantity is:$=L^{-1-2}=L^{-3}$ $[M^0 L^{-3} T^0]$Compare with $M^a L^b T^c$ $a=0, \;\; b=-3, \;\; c=0$

$R=\;\frac{R_2 R_1}{R_2-R_1}=\;\frac{4 \times 2}{2}=4 cm$

$U\;=\;\frac{1}{2} \times y \times (\;\text{ longitudinal strain }\;)^2$ $\sigma \;={| \;\frac{\;\text{ transverse strain }\;}{\;\text{ longitudinal strain }\;} |}$ $\;\text{ Longitudinal strain }\;=\;\;\frac{10^{-3}}{0.2}=5 \times 10^{-3}$ $U\;=\;\frac{1}{2} \times 2 \times 10^{11} \times 25 \times 10^{-6}$ $\;=25 \times 10^5 {\text{J}} / {\text{m}}^3$

$\;\Delta p=h \rho g+\;\frac{2 s}{r}$ $\;=0.2 \times 10^3 \times 10+\;\frac{2 \times 0.095}{10^{-3}}$ $\;=2000+190=2190 Pa$