Vital Capacity
Trending Questions
What is the vital capacity of our lungs?
Inspiratory reserve volume plus expiratory reserve volume
Inspiratory reserve volume plus tidal volume
Total lung capacity minus expiratory reserve volume
Total lung capacity minus residual volume
- Inspiratory Reserve Volume (IRV) + Expiratory Reserve Volume (ERV)+Tidal Volume (TV)+Residual Volume (RV)
- IRV+RV+ERV
- IRV+TV+ERV
- TV+RV+ERV
Respiratory capacities | Respiratory volumes |
(i) Residual volume | 2500 mL |
(ii) Vital capacity | 3500 mL |
(iii) Inspiratory reserve volume | 1200 mL |
(iv) Inspiratory capacity | 4500 mL |
Which one of the following is the correct matching of two capacities and volumes?
- (ii) 2500 mL, (iii) 4500 mL
- (iii) 1200 mL, (iv) 2500 mL
- (iv) 3500 mL, (i) 1200 mL
- (i) 4500 mL, (ii) 3500 mL
- 7 hours
- 14 hours
- Less than 7 hours
- More than 7 hours
One end of a copper rod of length 1.0 m and area of cross-section 10−3 is immersed in boiling water and the other end in ice. If the coefficient of thermal conductivity of copper is 92cal/m−s−∘C and the latent heat of ice is 8×104cal/kg, then the amount of ice which will melt in one minute is
9.2×10−3kg
8×10−3kg
6.9×10−3kg
5.4×10−3kg
- 5 W/mK
- 0.05 W/mK
- 0.5 W/mK
- 50 W/mK
- r1r2(r1−r2)
- (r2−r1)
- (r2−r1)(r1r2)
- In (r2r1)
- 6∘C
- 12∘C
- 18∘C
- 24∘C
- vital capacity
- total lung capacity
- functional residual capacity
- none of the above
Four rods of identical cross-sectional area and made from the same metal form the sides of square. The temperature of two diagonally opposite points is T and √2 T respective in the steady state. Assuming that only heat conduction takes place, what will be the temperature difference between other two points
√2+12T
2√2+1T
0
None of these
- 10 K
- 20 K
- 30 K
- 40 K
- 0.05 m
- 0.1 m
- 0.2 m
- 0.5 m
The maximum volume of air that can be released from the lungs by forceful expiration after deepest inspiration is called the
total lung capacity
vital capacity
tidal volume
ventilation rate
- K1T1+K2T2d1+d2
- K1T1d2+K2T2d1K1d2+K2d1
- (K1d1+K2d2)T1T2T1+T2
- K1d1T1+K2d2T2K1d1+K2d2
- 16.5∘C
- 22.5∘C
- 28.5∘C
- 32.5∘C
Two slabs A and B of different materials but of the same thickness are joined as shown in the figure. The thermal conductivities of A and B are k1 and k2 respectively. The thermal conductivity of the composite slab will be
(k1+k2)
12(k1+k2)
√k1k2
2k1k2(k1+k2)
Which of these is the sum of the other three?
Tidal volume
Expiratory reserve volume
Vital capacity
Inspiratory reserve volume
- heat flown trough A and E slabs are same
- heat flow through slab e is maximum
- temperature difference across slab E is smallest
- heat flow through C = heat flow through B + heat flow through D
- 1500ml
- 2000ml
- 2300ml
- 4300ml
1. Steady- state one-dimensional conduction.
2. Constant value of thermal conductivity
3. The bounding surfaces are isothermal.
4. Temperature gradient is constant.
of these statements:-
- 1 , 2 , and 3 are correct
- 2 and 4 are correct
- 2 , 3 , and 4 are correct
- 1, 3 and 4 are correct
- 1:24
- 1:8
- 1:4
- 1:6
- higher
- lower
- same
- 18πka
- 8a2πk
- 14πka
- 12πka
[Neglect heat capacity of steel rod and assume no heat is lost to surroundings. Also, take specific heat of water =S J/kg∘C, conductivity of steel ksteel=k W/m∘C]
- Skln2
- S2kln2
- 2Skln2
- 2kSln2
- Yes
- No
- K1>K2
- K1=K2
- K1=0
- K1<K2
(Neglect any heat loss to the surroundings)