You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Pressure

A thin tube o...

Question

A thin tube of uniform cross-section is sealed at both ends. It lies horizontally. The middle $5 c m$ contains Hg and two equal ends contain air at the same pressure $P_{0}$ . When the tube is held at an angle of $60^{o}$ with the vertical, the length of the air column above and below the Hg are $46 c m$ and $44.5 c m$ . Calculate pressure $P_{0}$ in cm of Hg. Assume temperature of the system to be constant.

55 c m

of Hg

No worries! We‘ve got your back. Try BYJU‘S free classes today!

65 c m

of Hg

No worries! We‘ve got your back. Try BYJU‘S free classes today!

70.4 c m

of Hg

No worries! We‘ve got your back. Try BYJU‘S free classes today!

75.4 c m

of Hg

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $75.4 c m$ of Hg
As the length of the tube remains constant, thus
$2 L + 5 = 46 + 5 + 44.5$
or
$L = 45.25 c m$
In case (ii) $P_{A} = P_{B} + 5 cos 60^{o}$
or $P_{A} - P_{B} = 2.5 c m$ of Hg
$P_{A} (44.5) = P_{B} (46)$
or $(\frac{46}{44.5} - 1) P_{B} = 2.5$
or $P_{B} = \frac{2.5 \times 44.5}{1.5}$
$P_{0} (44.25) = P_{B} \times 46$
Thus $P_{0} = P_{B} \frac{46}{45.25} = \frac{2.5}{1.5} \times \frac{44.5 \times 46}{45.25}$
$= 75.4 c m$ of Hg

Suggest Corrections

Similar questions

Q. A thin tube of uniform cross-section is sealed at both ends. It lies horizontally, the middle

5 c m

containing

H g

and the two equal ends containing air at the same pressure

P_{0}

. When the tube is held at an angle

60^{\circ}

with the vertical, the lengths of the air column above and below the mercury are

46 c m

and

44.5 c m

respectively. Calculate pressure

P_{0}

in cm of

H g

. (The temperature of the system is kept at

30^{\circ} C

A thin tube of uniform cross section is sealed at both ends. It lies horizontally, the middle 5 cm containing mercury and the parts on its two sides containing air at the same pressure P. When the tube is held at an angle of $60^{\circ}$ with the vertical, the length of the air column above and below the mercury pellet are 46cm and 44.5cm respectively. Calculate the pressure P in centimeters of mercury. The temperature of the system is kept at $30^{\circ} C$ .

A thin tube of uniform cross-section is sealed at both ends. It lies horizontally, the middle 5 cm containing mercury and the two equal ends containing air at the same pressure P $_{o}$ . When the tube is held at an angle 60 $^{o}$ with the vertical, the lengths of the air column above and below the mercury are 46 and 44.5 cm respectively. The pressure P $_{o}$ in cm of Hg is: (The temperature of the system is left at 30 K.)

A glass tube sealed at both ends is 1m long. It lies horizontally with the middle 10 cm containing Hg. The two ends of the tube equal in length contain air at 27 $^{0}$ C and pressure 76 cm of Hg. The temperature at one end is kept 0 $^{0}$ C and at the other end it is 127 $^{0}$ C Neglect the change in length of Hg column. Then the change in length on two sides is

Q. A glass tube sealed at both ends is

1

m long. It lies horizontally with the middle

10

cm containing Hg. The two ends of the tube, equal in length, contain air at

27^{o} C

and pressure

76

cm of Hg. The temperature at one end is kept

0^{o} C

and at the other end it is

127^{o} C

. Neglect the change in length of Hg column. Then the change in length on two sides is

Join BYJU'S Learning Program

The correct option is D 75.4cm of HgAs the length of the tube remains constant, thus2L+5=46+5+44.5 orL=45.25cmIn case (ii) PA=PB+5cos60oor PA−PB=2.5cm of HgPA(44.5)=PB(46)or (4644.5−1)PB=2.5or PB=2.5×44.51.5P0(44.25)=PB×46Thus P0=PB4645.25=2.51.5×44.5×4645.25=75.4cm of Hg