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  An object dropped from a leaf false with a constant acceleration of 10 m/s 2 . Find its Speed 2 second after it was dropped.                            Solution: Given: Initial speed, u = 0 u = 0 (object is dropped) Acceleration due to gravity, a = 10   m/s 2 a = 10 \, \text{m/s}^2 Time, t = 2 \, \text{s} t  = 2 s Formula used: v = u + a t Calculation: v = 0 + ( 10 × 2 ) = 20   m/s Answer: The speed of the object after 2 seconds is 20 m/s (downwards) .

                                A train travels 20 km at a uniform speed of 60 km/h and the next 20 km at a uniform speed of 80 km/hr. calculate its average speed. Given: First distance d 1 = 20   km d_1 = 20 \, \text{km} , speed v 1 = 60   km/h v_1 = 60 \, \text{km/h} Second distance d 2 = 20   km d_2 = 20 \, \text{km} , speed v 2 = 80   km/h v_2 = 80 \, \text{km/h} Step 1: Time taken for each part t 1 = 20 60 = 1 3  h t_1 = \frac{20}{60} = \frac{1}{3} \text{ h} t 2 = 20 80 = 1 4  h t_2 = \frac{20}{80} = \frac{1}{4} \text{ h} Step 2: Total distance and total time Total distance = 20 + 20 = 40  km \text{Total distance} = 20 + 20 = 40 \text{ km} Total time = 1 3 + 1 4 = 4 + 3 12 = 7 12  h \text{Total time} = \frac{1}{3} + \frac{1}{4} = \frac{4 + 3}{12} = \frac{7}{12} \text{ h} Step 3: Average speed Average speed = Total distance Total time = 40 7 / 12 =...

    A boy is running on a straight road. He runs 500 m towards north in 2 minutes 10 seconds and then turns back and runs 200 m in 1 minute. Calculate             i) His average speed and magnitude of average velocity during first 2 minutes 10 seconds and             ii) His average speed and magnitude of average velocity during the whole journey. Solution: Given: Distance covered towards north  = 500 m Time taken = 2 minutes 10 seconds = 130 s Then he runs back 200 m in 1 minute = 60 s (i) First 2 minutes 10 seconds This is only the first part of the motion. Distance covered = 500 m Time taken = 130 s (a) Average Speed :- Average Speed = Total Distance / Total Time = 500 / 130 = 3.85 m/s (b) Magnitude of Average Velocity :-      Displacement = 500 m towards north       Average velocity = Displacement / Time = 500 / 130 = 3.85 m/s (ii) Whole journey Total distance covered = 5...

                                   A taxi driver noted the reading on the odometer fitted in the vehicle as 1052 km when he started the journey. After 30 minutes drive, he noted that the odometer reading was 1088 km. find the average speed of the taxi. Solution: Given: Initial odometer reading = 1052 km Final odometer reading = 1088 km Time taken = 30 minutes = 0.5 hour Distance travelled: 1088 − 1052 = 36  km 1088 - 1052 = 36 \text{ km} Average speed: Average speed = Distance Time = 36 0.5 = 72  km/h \text{Average speed} = \frac{\text{Distance}}{\text{Time}} = \frac{36}{0.5} = 72 \text{ km/h} Average speed of the taxi = 72 km/h

  Heredity – Important MCQs (Class 10) 1–20: Basics of Heredity The transmission of traits from parents to offspring is called A) Variation B) Evolution C) Heredity D) Adaptation Ans: C The scientific study of heredity is known as A) Ecology B) Genetics C) Taxonomy D) Embryology Ans: B Who is known as the Father of Genetics? A) Darwin B) Lamarck C) Mendel D) Morgan Ans: C Mendel conducted experiments on A) Pea plant B) Wheat C) Rice D) Maize Ans: A The alternative forms of a gene are called A) Chromosomes B) Alleles C) Traits D) DNA Ans: B The gene controlling a single trait is called A) Polygenic B) Monogenic C) Digenic D) Multigenic Ans: B A dominant trait is expressed when A) Present only in homozygous condition B) Present in heterozygous condition...

 Numerical: A wave moves a distance of 8 m in 0.05 s. Find the velocity of the wave. Solution: Given: Distance d  =  8   m Time t  =  0.05   s Velocity of the wave is given by: v = d t v = \frac{d}{t} ​ v = 8 0.05 = 160   m/s v = \frac{8}{0.05} = 160 \,\text{m/s} Velocity of the wave = 160 m/s

  Numerical: A sound has 13 crests and 15 troughs in 3 seconds. When the second crest is produced the first is 2cm away from the source? Calculate a) The wavelength, b) The frequency, c) The wave speed Solution: Given data: Number of crests = 13 Number of troughs = 15 Time = 3 s Step 1: Find number of complete waves One complete wave = 1 crest + 1 trough Total wave parts = 13 + 15 = 28 So, number of complete waves n = 28 2 = 14  waves n = \frac{28}{2} = 14 \text{ waves} (a) Wavelength When the second crest is produced, the first crest is 2 cm away from the source. Distance between two successive crests = wavelength. λ = 2  cm = 0.02  m \lambda = 2 \text{ cm} = 0.02 \text{ m} Wavelength = 0.02 m (b) Frequency f = Number of waves Time f = \frac{\text{Number of waves}}{\text{Time}} ​ f = 14 3 ≈ 4.67  Hz f = \frac{14}{3} \approx 4.67 \text{ Hz} Frequency ≈ 4.67 Hz (c) Wave speed v = f λ v = f \lambda v = 4.67 × 0.02 ≈ 0.093  m/s v = 4...

  Numerical: Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium. Solution: Given: Frequency f  =  220   Hz Speed of sound v  =  440   m/s Wavelength is given by: λ = v f \lambda = \frac{v}{f} ​ λ = 440 220 = 2   m \lambda = \frac{440}{220} = 2 \,\text{m} Wavelength of the sound wave = 2 m

  Numerical: A source of wave produces 40 crests and 40 troughs in 0.4 second. Find the frequency of the wave. Solution: In one complete wave, there is one crest and one trough . Given: Number of crests = 40 Number of troughs = 40 So, number of complete waves = 40 Time taken = 0.4 s Frequency is given by: f = Number of waves Time f = \frac{\text{Number of waves}}{\text{Time}} ​ f = 40 0.4 = 100  Hz f = \frac{40}{0.4} = 100 \text{ Hz} Frequency of the wave = 100 Hz

  Numerical:  Find the frequency of a wave whose time period is 0.002 second. Solution: Given time period, T  =  0.002  s Frequency is given by: f = 1 T f = \frac{1}{T} ​ f = 1 0.002 = 500  Hz f = \frac{1}{0.002} = 500 \text{ Hz} Frequency of the wave = 500 Hz

  Heredity class 10 Notes Introduction Reproductive processes produce new individuals that are similar but show some variations, even in asexual reproduction. Sexual reproduction increases the chances of successful variations, which is why organisms like humans show more visible differences than plants like sugarcane. This chapter studies how variations are produced and how they are inherited from one generation to the next. Accumulation of Variation During Reproduction Accumulation of variation means the gradual build-up of small differences (variations) in organisms over many generations during the process of reproduction.  These variations are passed from parents to offspring and get added generation after generation. How does variation arise during reproduction? 1. Asexual Reproduction Only one parent is involved. Offspring are almost identical to the parent. Small variations occur due to: Errors in DNA copying Environmental effects Variations are fe...