Mastering Physics Solutions Chapter 30 Quantum Physics

Mastering Physics Solutions Chapter 30 Quantum Physics

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Chapter 30 Quantum Physics Q.1P
CE Predict/Explain The blackbody spectrum of blackbody A peaks at a longer wavelength than that of blackbody B. (a) Is the temperature of blackbody A higher than or lower than the temperature of blackbody B? (b) Choose the best explanation from among the following:

• Blackbody A has the higher temperature because the higher the temperaturethe longer the wavelength.
• Blackbody B has the highertemperature because an increase in temperature means an increase in frequency, which corresponds to a decrease in wavelength.

Solution:

Mastering Physics Solutions Chapter 29 Relativity

Mastering Physics Solutions Chapter 29 Relativity

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Chapter 29 Relativity Q.2P
Albert is piloting his spaceship, heading east with a speed of 0.90c. Albert’s ship sends a light beam in the forward (east-ward) direction, which travels away from his ship at a speed c. Meanwhile, Isaac is piloting his ship in the westward direction, also at 0.90c, toward Albert’s ship. With what speed does Isaac see?
Solution:
According to second postulate of special theory of relativity, the speed of light in vacuum, , is same in all inertial frames of reference, independent of the motion of the source or the receiver. The speed of Albert’s light beam observed by Isaac is c
Chapter 29 Relativity Q.3CQ
How would velocities add if the speed of light were infinitely large? Justify your answer by considering Equation 29–4.
Solution:

RS Aggarwal Class 10 Solutions Volume and Surface Areas of Solids

RS Aggarwal Class 10 Solutions Volume and Surface Areas of Solids

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Question 1:

Radius of the cylinder = 14 m
And its height = 3 m
Radius of cone = 14 m
And its height = 10.5 m
Let l be the slant height

Curved surface area of tent
= (curved area of cylinder + curved surface area of cone)

Hence, the curved surface area of the tent = 1034
Cost of canvas = Rs.(1034 × 80) = Rs. 82720

RS Aggarwal Class 10 Solutions Areas of Circle, Sector and Segment

RS Aggarwal Class 10 Solutions Areas of Circle, Sector and Segment

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APlusTopper.com Provides RS Aggarwal Class 10 Solutions with Free PDF download option. All questions are solved by expert Mathematic Teachers as per NCERT (CBSE) guidelines.

Question 2:
Circumference of circle = 2πr = 39.6 cm

RS Aggarwal Class 10 Solutions Perimeter and Areas of Plane Figures

RS Aggarwal Class 10 Solutions Perimeter and Areas of Plane Figures

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APlusTopper.com Provides RS Aggarwal Class 10 Solutions with Free PDF download option. All questions are solved by expert Mathematic Teachers as per NCERT (CBSE) guidelines.
Question 2:
If the cost of sowing the field is Rs. 58, then area = 10000 m²
If the cost of sowing is Re. 1, area =  [latex]\frac { 10000 }{ 58 }  [/latex] m²
If the cost of sowing is Rs. 783, area =  [latex](\frac { 10000 }{ 58 } \times 783)   [/latex] m²
Area of the field = 135000 m²
Let the attitude of the field be x meters
Then, Base = 3x meter
Area of the field =

Hence, the altitude = 300m and the base = 900 m

RS Aggarwal Class 10 Solutions Co-ordinate Geometry

RS Aggarwal Class 10 Solutions Co-ordinate Geometry

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APlusTopper.com Provides RS Aggarwal Class 10 Solutions with Free PDF download option. All questions are solved by expert Mathematic Teachers as per NCERT (CBSE) guidelines.

Question 3:
The given points are A(a, -1) and B(5,3).

RS Aggarwal Class 10 Solutions Probability

RS Aggarwal Class 10 Solutions Probability

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APlusTopper.com Provides RS Aggarwal Class 10 Solutions with Free PDF download option. All questions are solved by expert Mathematic Teachers as per NCERT (CBSE) guidelines.
Question 1:
(i) The probability of an impossible event is 0
(ii) The probability of a sure event is 1
(iii) For any event E, P(E) + P(not E) = 1
(iv) The probability of a possible but not a sure event lies between 0 and 1
(v) The sum of probabilities of all the outcomes of an experiment is 1
Question 2:
When a coin is tossed, all possible outcomes are either H or T
Total number of possible outcomes = 2
The favorable outcome is T
Number of favorable outcomes = 1
∴ P(getting a T) =  [latex]P(T)=\frac { Number\quad of\quad favorable\quad outcomes }{ Total\quad number\quad of\quad possible\quad outcomes } =\frac { 1 }{ 2 }   [/latex]

RS Aggarwal Class 10 Solutions Height and Distance

RS Aggarwal Class 10 Solutions Height and Distance

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APlusTopper.com Provides RS Aggarwal Class 10 Solutions with Free PDF download option. All questions are solved by expert Mathematic Teachers as per NCERT (CBSE) guidelines.
Question 1:
Let AB be the tower standing on a level ground and O be the position of the observer. Then OA = 20 m and ∠OAB = 90° and ∠AOB = 60°

Let AB = h meters
From the right ∆OAB, we have

Hence the height of the tower is [latex]20\sqrt { 3 } m=34.64m    [/latex]

RS Aggarwal Class 10 Solutions Constructions

RS Aggarwal Class 10 Solutions Constructions

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APlusTopper.com Provides RS Aggarwal Class 10 Solutions with Free PDF download option. All questions are solved by expert Mathematic Teachers as per NCERT (CBSE) guidelines.
Question 1:

Steps of construction:
Step 1 : Draw a line segment AB = 6.5 cm
Step 2: Draw a ray AX making ∠ BAX.
Step 3: Along AX mark (4+7) = 11 points
A₁, A₂, A₃, A₄, A₅, A₆, A₇, A₈, A₉, A₁₀, A₁₁, such that
AA₁ = A₁A₂
Step 4: Join A₁₁ and B.
Step 5: Through A₄ draw a line parallel to A₁₁ B meeting AB at C.
Therefore, C is the point on AB, which divides AB in the ratio 4 : 7
On measuring,
AC = 2.4 cm
CB = 4.1 cm