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स्वागत गीत : अथ स्वागतम् शुभ स्वागतम् (Ath Swagatam Shubh Swagatam)

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स्वागत गीत : अथ स्वागतम् शुभ स्वागतम् (Ath Swagatam Shubh Swagatam)  अथ स्वागतं शुभ स्वागतम्  स्वागतम् । अथ स्वागतं शुभ स्वागतम् । आनंद मंगल मंगलम् । नित प्रियं भारत भारतम् ॥ ध्रु.॥ नित्य निरंतरता नवता मानवता समता ममता सारथि साथ मनोरथ का जो अनिवार नहीं थमता संकल्प अविजित अभिमतम् ॥ १॥ आनंद मंगल मंगलम् । नित प्रियं भारत भारतम् । अथ स्वागतं शुभ स्वागतम् ॥ कुसुमित नई कामनाएँ सुरभित नई साधनाएँ मैत्रीमात क्रीडांगन में प्रमुदित बन्धु भावनाएँ शाश्वत सुविकसित इति शुभम् ॥ २॥ आनंद मंगल मंगलम् । नित प्रियं भारत भारतम् । अथ स्वागतं शुभ स्वागतम् ॥

Ncert Solution CBSE Class 11 Physics chapter 9 MECHANICAL PROPERTIES OF FLUIDS - Param Himalaya

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9.1 Explain why (a) The blood pressure in humans is greater at the feet than at the brain. (b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km. (c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area. Solution:  (a) $P = h \rho g$. The height (h) of the blood column in the human body is more at the feet than at the brain. Therefore , blood pressure in humans is greater at the feet than at the brain. (b) Atmospheric pressure $P = h \rho g$ where $\rho$ is the density of air. As we go up , the density of air decrease rapidly and at a height of about 6 km , it decreases to nearly half of its value at the sea level. Due to this reason , atmospheric pressure at a height of about 6 Km decreases to nearly half of its value at the sea level. (C) When force is applied on a liquid , hydrostatic pressure is transmitted equally in all directions in ...

Ncert Solution CBSE Class 11 Chapter 8 MECHANICAL PROPERTIES OF SOLIDS - Param Himalaya

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Ncert Solution CBSE Class 11 Chapter 8 MECHANICAL PROPERTIES OF SOLIDS - Param Himalaya  8.1 A steel wire of length 4.7 m and cross-sectional area $3.0 \times 10^{-5}m^{2}$ stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of $4.0 \times 10^{-5}m^{2}$ under a given load. What is the ratio of the Young’s modulus of steel to that of copper? Solution :  $$Y= \frac{F/A}{\Delta L/L}$$ Or $$Y=\frac{FL}{\Delta L A}$$ Here F (Load) and $\Delta L$ (elongation) are the same in the two cases :  For steel :  $$Y_{s}=\frac{F \times4.7}{3.0 \times 10^{-5}\times \Delta L}$$ For copper : $$Y_{c}=\frac{F \times3.5}{4.0 \times 10^{-5}\times \Delta L}$$ $$\therefore \frac{Y_{s}}{Y_{C}} = \frac{4.7\times 4.0 \times 10^{-5}}{3.0\times 3.0 \times 10^{-5}} = 1.79$$ 8.2 Figure 8.9 shows the strain-stress curve for a given material. What are (a) Young’s  modulus and (b) approximate yield strength for this material? Solution:  (a) From the stress...

Ncert Solution CBSE Class 11 Physics Chapter 7 Gravitation - Param Himalaya

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Ncert Solution CBSE Class 11 Physics Chapter 7 Gravitation - Param Himalaya 7.1 Answer the following : (a) You can shield a charge from electrical forces by putting it inside a hollow conductor.Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means ? Solution: You cannot shield a body from the gravitational influence of nearby matter by any means. It is because the gravitational force on a body due to nearby matter is not altered due to the presence of other bodies. In other words , gravitational field cannot be shielded by any means. (b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity ? Solution: Yes , If the size of the spaceship orbiting earth is very large , an astronaut in it can detect the gravity. (c) If you compare the gravitational force on the earth due to the sun to...