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VIDYANAND JHA
I’m an accidental blogger and a microbiologist by education. I have dedicated my past 5+ years of learning to the field of life science. helping numerous individuals and providing useful educational content around the globe to help them grow in the field of biology. Helping others grow has always been my biggest driving force, and this story will tell you about the important time period of this journey.
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Respiratory Quotient & Factors Affecting Respiration MCQ Test | Topic-Wise NEET Biology
Practice topic-wise NEET Biology MCQs on Respiratory Quotient (RQ) and Factors Affecting Respiration based on NCERT concepts. Test your unde...
Fermentation & Amphibolic Pathway – Alcoholic, Lactic Acid MCQ Test | Topic-Wise NEET Biology
Practice topic-wise NEET Biology MCQs on Fermentation and Amphibolic Pathway, covering alcoholic fermentation, lactic acid fermentation, and...
Electron Transport System & Oxidative Phosphorylation MCQ Test | Topic-Wise NEET Biology
Practice topic-wise NEET Biology MCQs on the Electron Transport System (ETS) and Oxidative Phosphorylation based on NCERT concepts. Test you...
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Plasmolysis: Definition, Types, Forms, Examples and Significance
Learn what plasmolysis is, how it works in plant cells, and why it matters in biology with clear definitions and examples.
Mucous Membrane (Mucosa): Structure, Locations, Functions and Disorders
Learn about the structure, locations, and functions of mucous membranes. Discover common disorders and how they protect the body.
Bone Cells: Types, Structure, Functions, Remodeling and Disorders
Learn about the three main types of bone cells—osteoblasts, osteocytes, and osteoclasts—their roles in bone formation, maintenance, and remodeling.
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Practice questions contributed by this author
How much heat is required to vaporize 0.25kg of ethanol at 78∘C? (Latent heat of vaporization of ethanol = 8.5×105J kg−1
Given: m = 0.25kg, Lv = 8.5×105J kg−1. Q = mLv = 0.25×8.5×105 = 212500J = 212.5kJ. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 212.5 kJ. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
A gas at 2.2atm and 57∘C occupies 5.5L. If the volume is reduced to 3.3L and temperature increased to 107∘C, what is the
Given: P1 = 2.2atm, T1 = 57∘C = 330K, V1 = 5.5L, V2 = 3.3L, T2 = 107∘C = 380K. P1V1T1 = P2V2T2. P2 = P1×V1V2×T2T1 = 2.2×5.53.3×380330. P2 = 2.2×1.6667×1.1515≈4.22atm.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
A gas at 2atm and 27∘C occupies 5L. What pressure is required to compress it to 2L while maintaining the temperature at
Given: P1 = 2atm, V1 = 5L, T1 = 27∘C = 300K, V2 = 2L, T2 = 327∘C = 600K. P1V1T1 = P2V2T2. P2 = P1×V1V2×T2T1 = 2×52×600300 = 2×2.5×2 = 10atm.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.