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A body of mass 0.5 kg travels in a straight line with velocity

[question] Question. A body of mass $0.5 \mathrm{~kg}$ travels in a straight line with velocity $v=a x^{\frac{3}{2}}$ where $a=5 \mathrm{~m}^{-\frac{1}{2}} \mathrm{~s}^{-1}$. What is the work done by the net force during its displacement from $x=0$ to $x=2$ m? [/question] [solution] solution: Mass of the body, m = 0.5 kg Velocity of the body is governed by the equation, $v=a x^{\frac{3}{2}}$ with $a=5 \mathrm{~m}^{\frac{-1}{2}} \mathrm{~s}^{-1}$ Initial velocity, $u($ at $x=0)=0$ Final velocity ...

The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 105 ms–1.

Question: The velocity associated with a proton moving in a potential difference of $1000 \mathrm{~V}$ is $4.37 \times 10^{5} \mathrm{~ms}^{-1}$. If the hockey ball of mass $0.1 \mathrm{~kg}$ is moving with this velocity, calculate the wavelength associated with this velocity. Solution: According to de Broglie's expression, $\lambda=\frac{\mathrm{h}}{m v}$ Substituting the values in the expression, $\lambda=\frac{6.626 \times 10^{-34} \mathrm{Js}}{(0.1 \mathrm{~kg})\left(4.37 \times 10^{5} \math...

If $\sin A=\frac{\mathbf{3}}{\mathbf{4}}$, calculate $\cos A$ and $\tan A$.

[question] Question. If $\sin A=\frac{\mathbf{3}}{\mathbf{4}}$, calculate $\cos A$ and $\tan A$. [/question] [solution] Solution: In figure, $\sin A=\frac{3}{4}$ $\Rightarrow \frac{B C}{A C}=\frac{3}{4}$ $\Rightarrow \mathrm{BC}=3 \mathrm{k}$ and $A C=4 k$ where k is the constant of proportionality. By Pythagoras Theorem, $\mathrm{AB}^{2}=\mathrm{AC}^{2}-\mathrm{BC}^{2}=(4 \mathrm{k})^{2}-(3 \mathrm{k})^{2}=7 \mathrm{k}^{2}$ $\Rightarrow \mathrm{AB}=\sqrt{7} \mathrm{k}$ So, $\cos A=\frac{A B}{A ...

If the velocity of the electron in Bohr’s first orbit is 2.19 × 106 ms–1,

Question: If the velocity of the electron in Bohr's first orbit is $2.19 \times 10^{6} \mathrm{~ms}^{-1}$, calculate the de Broglie wavelength associated with it. Solution: According to de Broglie's equation, $\lambda=\frac{h}{m v}$ Where, $\lambda=$ wavelength associated with the electron $h=$ Planck's constant $m=$ mass of electron $v=$ velocity of electron Substituting the values in the expression of $\lambda$ : $\lambda=\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.10939 \times 10^{-31} \...

Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules.

Question: Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron. Solution: From de Broglie's equation, $\lambda=\frac{\mathrm{h}}{m v}$ $v=\frac{\mathrm{h}}{m \lambda}$ Where, $v=$ velocity of particle (neutron) $h=$ Planck's constant $m=$ mass of particle (neutron) $\lambda=$ wavelength Substituting the values in the...

Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material

Question: Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is $1.6 \times 10^{6} \mathrm{~ms}^{-1}$, calculate de Broglie wavelength associated with this electron. Solution: From de Broglie's equation, $\lambda=\frac{\mathrm{h}}{m v}$ $\lambda=\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.10939 \times 10^{-...

Describe transpiration pull model of water transport in plants. What are the factors influencing transpiration? How is it useful to plants?

Question: Describe transpiration pull model of water transport in plants. What are the factors influencing transpiration? How is it useful to plants? Solution: In tall trees, water rises with the help of the transpirational pull generated by transpiration or loss of water from the stomatal pores of leaves. This is called the cohesion-tension model of water transport. During daytime, the water lost through transpiration (by the leaves to the surroundings) causes the guard cells and other epiderma...

The angles of a elevation of the top of a tower from two points

Question: The angles of a elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m. Solution: Let the tower be represented by AB in the figure. Let AB = h metres. $\therefore$ In right $\Delta A B C$, we have $\frac{\boldsymbol{A B}}{\boldsymbol{A C}}=\tan \theta$ $\Rightarrow \frac{\mathbf{h}}{\mathbf{9}}=\tan \theta$ ...(1) In right $\triangle \mathr...

A solid cylinder rolls up an inclined plane of angle of inclination 30°.

Question: A solid cylinder rolls up an inclined plane of angle of inclination $30^{\circ}$. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 $\mathrm{m} / \mathrm{s}$ (a) How far will the cylinder go up the plane? (b) How long will it take to return to the bottom? Solution: A solid cylinderrolling up an inclination is shown in the following figure. Initial velocity of the solid cylinder,v= 5 m/s Angle of inclination, $\theta=30^{\circ}$ Height reached by th...

A straight highway leads to the foot of a tower.

Question: A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60. Find the time taken by the car to reach the foot of the tower. Solution: Let PQ = h metres be the height of the tower. P is the top of the tower. PX is horizontal line through P. The first and second positions of the...

What role does root pressure play in water movement in plants?

Question: What role does root pressure play in water movement in plants? Solution: Root pressure is the positive pressure that develops in the roots of plants by the active absorption of nutrients from the soil. When the nutrients are actively absorbed by root hairs, water (along with minerals) increases the pressure in the xylem. This pressure pushes the water up to small heights. Root pressure can be observed experimentally by cutting the stem of a well-watered plant on a humid day. When the s...

How is the mycorrhizal association helpful in absorption of water and minerals in plants?

Question: How is the mycorrhizal association helpful in absorption of water and minerals in plants? Solution: Mycorrhiza is a symbiotic association of fungi with the root systems of some plants. The fungal hyphae either form a dense network around the young roots or they penetrate the cells of the roots. The large surface area of the fungal hyphae is helpful in increasing the absorption of water and minerals from the soil. In return, they get sugar and nitrogenous compounds from the host plants....

Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm.

Question: Calculate the wavelength for the emission transition if it starts from the orbithaving radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum. Solution: The radius of the $n^{\text {th }}$ orbit of hydrogen-like particles is given by, $r=\frac{0.529 n^{2}}{Z} A$ $r=\frac{52.9 n^{2}}{Z} \mathrm{pm}$ For radius $\left(r_{1}\right)=1.3225 \mathrm{~nm}$ $=1.32225 \times 10^{-9} \mathrm{~m}$ $=1322.25 \times 10^{-12} \mathrm{~m...

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground.

Question: A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60. After some time, the angle of elevation reduces to 30. Find the distance travelled by the balloon during the interval. Solution: From figure, we have $\angle \mathrm{POQ}=30^{\circ}$ is the angle of elevation for the first position of the balloon. Let $\mathrm{OQ}=\mathrm{y} \mathrm{m}$ W...

(a) With the help of well-labelled diagrams,

Question: (a) With the help of well-labelled diagrams, describe the process of plasmolysis in plants, giving appropriate examples. (b) Explain what will happen to a plant cell if it is kept in a solution having higher water potential. Solution: (a)Plasmolysis can be defined as the shrinkage of the cytoplasm of a plant cell, away from its cell wall and toward the centre. It occurs because of the movement of water from the intracellular space to the outer-cellular space. This happens when the plan...

The oxygen molecule has a mass of 5.30 × 10–26 kg and a moment of inertia of 1.94×10–46 kg m2

Question: The oxygen molecule has a mass of $5.30 \times 10^{-26} \mathrm{~kg}$ and a moment of inertia of $1.94 \times 10^{-46} \mathrm{~kg} \mathrm{~m}^{2}$ about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is $500 \mathrm{~m} / \mathrm{s}$ and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule. Solution: Mass of an oxygen mo...

As observed from the top of a 75m high lighthouse from the sea-level,

Question: As observed from the top of a 75m high lighthouse from the sea-level, the angles of depression of two ships are 30 and 45. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. Solution: In the figure, let AB represent the light house. $\therefore \quad \mathrm{AB}=75 \mathrm{~m}$Let the two ships be C and D such that angle of depression from A are $45^{\circ}$ and $30^{\circ}$ respectively. Now, in right $\triangle \mathrm...

What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?

Question: What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution? Solution: The water potential of pure water or a solution increases on the application of pressure values more than atmospheric pressure. For example: when water diffuses into a plant cell, it causes pressure to build up against the cell wall. This makes the cell wall turgid. This pressure is termed as pressure potential and has a positive value....

Briefly describe water potential. What are the factors affecting it?

Question: Briefly describe water potential. What are the factors affecting it? Solution: Water potential quantifies the tendency of water to move from one part to the other during various cellular processes such as diffusion, osmosis, etc. It is denoted by the Greek letter Psi or Ψ and is expressed in Pascals (Pa). The water potential of pure water is always taken as zero at standard temperature and pressure. Water potential $\left(\Psi_{w}\right)$ is expressed as the sum of solute potential $\l...

A hoop of radius 2 m weighs 100 kg.

Question: A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it? Solution: Radius of the hoop,r= 2 m Mass of the hoop,m= 100 kg Velocity of the hoop,v= 20 cm/s = 0.2 m/s Total energy of the hoop = Translational KE + Rotational KE $E_{\mathrm{T}}=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$ Moment of inertia of the hoop about its centre, $I=m r^{2}$ $E_{\mathrm{T}}=\frac{1}{2} m v^{2}+\frac...

From the top of a 7 m high building,

Question: From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60 and the angle of depression of its foot is 45. Determine the height of the tower. Solution: Let PQ = h metres be the height of the cable tower. AB = 7 metres is the height of the bulding $\angle \mathrm{PAR}=60^{\circ}$ is the angle of elevation of the top of the cable tower from the top of the building. $\angle \mathrm{RAQ}=45^{\circ}$ is the angle of depression of the foot of the cable tower...

Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented

Question: Emission transitions in the Paschen series end at orbit $n=3$ and start from orbit $n$ and can be represented as $v=3.29 \times 10^{15}(\mathrm{~Hz})\left[1 / 3^{2}-1 / n^{2}\right]$ Calculate the value ofnif the transition is observed at 1285 nm. Find the region of the spectrum. Solution: Wavelength of transition $=1285 \mathrm{~nm}$ $=1285 \times 10^{-9} \mathrm{~m}$ (Given) $v=3.29 \times 10^{15}\left(\frac{1}{3^{2}}-\frac{1}{n^{2}}\right)$ (Given) Since $v=\frac{c}{\lambda}$ $=\fra...

A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination.

Question: A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination. (a) Will it reach the bottom with the same speed in each case? (b) Will it take longer to roll down one plane than the other? (c) If so, which one and why? Solution: (a)Yes(b)Yes(c)On the smaller inclination (a)Mass of the sphere =m Height of the plane =h Velocity of the sphere at the bottom of the plane =v At the top of the plane, the total energy of the sphere = Potential ...

A TV tower stands vertically on a bank of a canal.

Question: A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60. From another point 20 m away this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30 (see fig.). Find the height of the tower and the width of the canal. Solution: Let $P Q=h$ metres be the height of the tower and $B Q=x$ metres be the width of the canal $\ang...

Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide.

Question: Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60 and 30, respectively. Find the height of the poles and the distances of the point from the poles. Solution: Let AB and CD be the towels P is the point between them. AB = h metres CD = h metres AP = x m CP = (80 x) m Now, in right $\triangle \mathrm{APB}$, We have $\frac{\mathbf{A B}}{\...

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