- The eigenvalues of the matrix
*A*, where^{-1}*A =*, are- 1 and 1/2
- 1 and 1.3
- 2 and 3
- 1/2 and 1/3

Answer:- 1/2 and 1/3

Eigenvalues of*A*are 2 and 3. So, eigenvalues of*A*are 1/2 and 1/3.^{-1} - The radius of the Earth is 6.37*10
^{6}m and the acceleration due to gravity at its surface is 9.81 m/s^{2}. A satellite is in circular orbit at a height of 35.9*10^{6}m above the earth’s surface. This orbit is inclined at 10.5 degrees to the equator. The velocity change needed to make the orbit equatorial is:- 561 m/s at 84.75 degrees to the initial direction
- 561 m/s at 95.25 degrees to the initial direction
- 281 m/s at 84.75 degrees to the initial direction
- 281 m/s at 95.25 degrees to the initial direction

Answer:-

- A piston-prop airplane having propeller efficiency,
*η*= 0.8 and weighing 73108 N could achieve maximum climb rate of 15 m/s at fight speed of 50 m/s. The excess Brake Power(BP) at the above flight condition will be_{p}- 1700 kW
- 2100 kW
- 1371 kW
- 6125 kW

Answer:- 1371 kW

Rate of Climb(R/C) = Excess Power / Weight

Therefore, Excess Power = 73108*15 = 1096620 W

Therefore, Excess Brake Power = Excess Power / Propeller Efficiency = 1370775 W = 1371 kW - An airplane model with a symmetric airfoil was tested in a wind tunnel. at angle of attack was estimated to be 0.08 and 0 respectively for elevator settings of 5 degrees up and 5 degrees down. The estimated value of the elevator control power of the model will be
- 0.07 per deg
- -1.065 per deg
- -0.008 per deg
- -0.762 per deg

Answer:-

- The lateral-directional characteristic equation for an airplane gave the following set of roots:
- An airplane is flying at an altitude of 10 km above the sea level. Outside air temperature and density at 10 km altitude are 223 K and 0.413 kg/m
^{3}respectively. The airspeed indicator of the airplane indicates a speed of 60 m/s. Density of air at sea level is 1.225 kg/m^{3}and value of gas constant R is 288 J/kg/K. The stagnation pressure(P_{0}) measured by the Pitot tube mpounted on the wing tip ofr the airplane will be of magnitude- 3.5*10
^{4}N/m^{2} - 2.0*10
^{4}N/m^{2} - 2.87*10
^{4}N/m^{2} - 0.6*10
^{4}N/m^{2}

Answer:- 2.87*10

^{4}N/m^{2}

The airspeed indicator shows equivalent airspeed. Therefore, sea level density is to be considered for velocity measurement.

*p = ρRT = 0.413*288*223 = 26525 N/m*^{2}

*0.5*ρ*_{s}*v^{2}= 0.5*1.225*60^{2}= 2205 N/m^{2}

Therefore,*P*_{0}= 26525+2205 = 28730 = 2.87*10^{4}N/m^{2} - 3.5*10