# Common Physical Biological Sciences Practice Test 2

Question – 1

1. The pH of a solution is 8.2 at 2500C. Calculate it’s [OH-].

• A. 5.8 M

• B. 1.58 x 10-6M

• C. 2.28 x10-4M

• D. 4.28 M

• E. 6.89 x 10-6M

pH + pOH = 14.0; pOH = 14.0 – pH
pOH = 14.0 – 8.2 = 5.8
pOH = 5.8
[OH]- = antilog [-5.8] = 1.58 x 10-6M

Question – 2

2. If a ball is thrown at an angle of 400 to the horizontal with a velocity of 25 m/s, what will be its time of flight?

• A. 3.3s

• B. 1.6s

• C. 3.9s

• D. 1.95s

• E. 1.65s

Time of flight consists of time of ascent and time of descent and is given as
Where u is the initial velocity;
g is the acceleration due to the gravity;

$\mathrm{??}$is the angle with which a projectile is being thrown from the horizontal x-axis
Since, u = 25 m/s, g = 9.8 m/s2 and

$\mathrm{??}$ = 400. Putting the values in the expressin above we get

Or,

$t={3.279}_{s}$Option a is the closest to the correct answer and so it is the correct answer option.

Question – 3

3. Tape worms belong to the phylum

• A. Cnidaria

• B. Platyhelminthes

• C. Mollusca

• D. Porifera

• E. Chordata

Tape worms are the class of Cestoda of the phylum Platyhelminthes. Platyhelminthes are free living, parasitic unsegmented worms.

Question – 4

4. A projectile is thrown at an angle of 600to the horizontal with a velocity of 17 m/s. What would be its angle after 2 seconds?

• A. ?? = tan-1 (-14.7/4.9)

• B. ?? = tan-1 (-4.9/14.7)

• C. ?? = tan-1 (-14.7/17)

• D. ?? = tan-1 (-4.9/17)

• E. ?? = tan-1 (-17/4.9)

• Answer Explanation:During the flight of the projectile horizontal velocity remain constant and vertical velocity changes. We can determine the angle by taking the tangent of vertical to horizontal velocity after 2s. Hence,
$\mathrm{??}=\mathrm{tan}\left(\frac{{v}_{y}}{{v}_{x}}\right)$Where

${v}_{x}$ is the horizontal velocity;

${v}_{y}$ is the vertical velocity

Since horizontal velocity remains constant

${v}_{x}$ = 17 m/s. Vertical velocity is given as

${v}_{y}={v}_{yo}???gt$
${v}_{yo}$ is the initial velocity;
Since, the projectile was thrown at an angle of 600 from x-axis, resolve the initial velocity in two components. One being the horizontal which is constant throughout the flight and the second is the vertical. It is given as 17 sin60 = 14.72 m/s;

g is the acceleration due to the gravity

t = 2 s

Therefore, vy = 14.72-9.8 x 2

Or, vy = -4.9.

Negative sign indicates that body is falling under gravity and the angle to be found is forming below the horizontal as shown.

Hence, d is the correct answer option

Question – 5

5. Write the equilibrium constant for the following reaction.SnO2(s) + 2CO(g) Sn(s) + 2CO2(g)

• A. Kc = [Sn][CO2]/[SnO2][CO]2

• B. Kc = [CO]2]/[CO2]2

• C. Kc = [Sn][CO2]2/[CO]2

• D. Kc = [CO2]2/[CO]2

• E. Kc = [CO2]2

The equilibrium constant for the above reaction is The equilibrium constant for the above reaction is Kc = [CO2]2/[CO]2The above reaction is an example for the heterogeneous equilibrium. In heterogeneous equilibrium, the concentrations of the solid and liquid are constant. Therefore, the equilibrium constant doesn’t contain the concentration of the solid and liquid.

Question – 6

6. If a liquid of density of 810 Kg/m3 and mass 0.950 Kg is kept in a vessel of negligible mass, what would be its volume in cm3?

• A. 1.17 x 10-3

• B. 11.7

• C. 117

• D. 1.17

• E. 1170

Density is given as

mass = 0.950 Kg
density = 810 Kg/m3
Putting the values in the expression above we get
volume =

$0.950810$ volume = 1.17 x 10-3m3
Since, 1m = 100 cm
Or,1m3 = 1003cm3
Or, 1m3 = 1000000cm3
Or, 1m3 = 106 cm3
Hence, converting volume in m3 to cm3 we get
volume = 1.17 x 10-3 x 106 cm3
Hence, e is the correct answer option.

Question – 7

7. Which of the following classes does the fungi Neurospora belong to?

• A. Phycomycetes

• B. Basidiomycetes

• C. Ascomycetes

• D. Deuteromycetes

• E. None of the above

Neurospora is an ascomycete mold with broadly spreading colonies. The name neurospora means ‘the nerve spore’ which refers to the axon like spores. Neurospora is widely used as model organism in genetic and biochemical work because of its quick reproductive nature.

Question – 8

8. What is the

$\frac{g???cm}{s}$equivalent of 156.7

$\frac{kg???m}{s}$?

• A. 1.567 x 105

• B. 1.567 x 10 6

• C. 1.567 x 10 7

• D. 1.567 x 10 3

• E. 1.567 x 104

Since 1 m = 102 cm and 1 Kg = 103 g. Therefore
156.7
$\frac{kg???m}{s}$ = 156.7 x 103 x 102

$\frac{g???cm}{s}$Or, 156.7

$\frac{kg???m}{s}$ = 15670000

$\frac{g???cm}{s}$Or, 156.7

$\frac{kg???m}{s}$ = 1.567 x 10 7

$\frac{g???cm}{s}$Hence, c is the correct answer option.

Question – 9

• A. Arthropods

• B. Molluscs

• C. Echinoderms

• D. Annelids

• E. Hemichordates

Madreporite is a porous structure by which the echinoderms filter sea water into their vascular system. It appear as bright orange dot on the aboral surface of star fish.

Question – 10

10. Which of the following is a dimensionally correct representation for Planck’s constant?

• A. ML2T-2

• B. MLT-2

• C. ML-1T2

• D. ML2T-1

• E. ML-1T-1

According to Planck’s law E = h??
Where
E is the energy of the photon;
h is the Planck’s constant;
?? is the frequency of the photon;
Dimensional formula for energy is ML2T-2and that of frequency is T-1Therefore, dimensional formula for Planck’s constant becomes
ML2T2 = h x T-1
Or, h = ML2T-1
Hence, d is the correct answer option.

Question – 11

11. A gas expands against the external pressure of 5 atm from a volume of 8L to 13 L. How much work is done?

• A. +40 atm.L

• B. -65 atm.L

• C. +25 atm.L

• D. -25 atm.L

• E. -40 atm.L

Work W = P x ?V = 5 atm x (13L – 8L) = 25 atm.L The work has been done by the system. Therefore, -W = P x ?V = -25 atm.L

Question – 12

12. Which of the following vegetative organs is a modified stem?

• A. Corm

• B. Rhizome

• C. Stolen

• D. Bulb

• E. All of the above

Corm, rhizome, stolen and bulb are the modified stems which are used in vegetative propagation. Vegetative propagation is a type of asexual reproduction in which a bud grows and develops into a new plant.

Question – 13

13. Compound A reacts by first order kinetics. The rate constant of the reaction is 0.45 sec-1. Calculate the half life of the compound A in the reaction.

• A. 4.62 seconds

• B. 3.08 seconds

• C. 1.54 seconds

• D. 2.25 seconds

• E. 0.9 seconds

Half life period of the first order reaction can be calculated using the following formula, which is derived from the first order rate law. t1/2 = 0.693/k1seconds. Therefore, t1/2 = 0.693/0.45 sec-1 = 1.54 seconds.

Question – 14

14. Which of the following is dimensionally incorrect?

• A. Force – MLT-2

• B. Pressure – ML-1T-2

• C. Volume – L3

• D. Intensity – ML1T-3

• E. Power – ML2T-3

Force = mass x acceleration
Dimensional formula of mass is M and acceleration is LT-2.Therefor,
Force =
$M??\frac{L}{{T}^{2}}$Or, Force = MLT-2
So, a is correct.
Pressure =

$Pressure=\frac{Force}{Area}$Dimensional formula of force is MLT-2 and that of area is L2
Pressure =

$\frac{ML{T}^{???2}}{{L}^{2}}$Or, Pressure = ML-1T-2
Therefore, b is correct.
SI unit of volume is m3. Hence, dimensional formula of volume becomes L3 . Therefore, c is correct as well.
Power is work done per unit time. Power =

$\frac{work}{time}$Or, Power =

$\frac{M{L}^{2}{T}^{???2}}{T}$Power = ML2T-3
So, e is correct as well.
Intensity is given as I =

$\frac{Power}{Area}$SI unit of intensity is

$\frac{watt}{{m}^{2}}$Or, Intensity =

$\frac{M{L}^{2}{T}^{???3}}{{L}^{2}}$Or, Intensity = MT-3
Hence, d is incorrect and so it is the right answer option.

Question – 15

15. Trichomonas is an example of

• B. Parabasalids

• C. Englenozoa

• D. Englenozoa

• E. None of the above