Properties of Matter 4 Question 2
4. A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance $d$ of $1.2 m$ from the person. In the following, state of the lift’s motion is given in List I and the distance where the water jet hits the floor of the lift is given in List II. Match the statements from List I with those in List II and select the correct answer using the code given below the lists.
(2014 Adv.)
| List I | List II | ||
|---|---|---|---|
| A | Lift is accelerating vertically up. | $p$ | $d=1.2 m$ |
| B | Lift is accelerating vertically down with an acceleration less than the gravitational acceleration. |
$q$ | $d>1.2 m$ |
| C | Lift is moving vertically up with constant speed. |
$r$ | $d<1.2 m$ |
| $s$ | No water leaks out of D |
Codes
(a) A-q, B-r, C-q, D-s
(b) A-q, B-r, C-p, D-s
(c) A-p, B-p, C-p, D-s
(d) A-q, B-r, C-p, D-p
Show Answer
Answer:
Correct Answer: 4. (d)
Solution:
- Let $R$ be the radius of the meniscus formed with a contact angle $\theta$. By geometry, this radius makes an angle $\theta+\frac{\alpha}{2}$ with the horizontal and,
$$ \cos \left(\theta+\frac{\alpha}{2}\right)=b / R $$
Let $P _0$ be the atmospheric pressure and $P _1$ be the pressure just below the meniscus. Excess pressure on the concave side of meniscus of radius $R$ is,
$$ P _0-P _1=2 S / R $$
The hydrostatic pressure gives,
$$ P _0-P _1=h \rho g $$
Eliminate $\left(P _0-P _1\right)$ from second and third equations and substitute $R$ from first equation to get,
$$ h=\frac{2 S}{\rho g R}=\frac{2 S}{b \rho g} \cos \left(\theta+\frac{\alpha}{2}\right) $$
